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Class 

Book 


Copyright N° 


COPYRIGHT DEPOSIT. 






























WORKS OF 


LIEUT.-COLONEL G. J. FIEBEGER 

PUBLISHED BY 

JOHN WILEY & SONS. 


A Text-book on Field Fortification. 

Small 8vo, xii + 166 pages, 27 colored maps. C loth, 
$2.00 net. 

Civil Engineering. 

A Text-book for a Short Course. 8vo, xiv-f-573 
pages, 180 figures. Cloth, $5.00 net. 








CIVIL ENGINEERING 


A TEXT-BOOK FOR A SHORT COURSE 


BY 

Lieut.-Col. G. J. FIEBEGER, U. S. Army 

Professor of Engineering, U. S. Military Academy 
Member of American Society of Civil Engineers 


FIRST EDITION 
FIRST THOUSAND 


NEW YORK 

JOHN WILEY & SONS 
London : CHAPMAN & HALL, Limited 


TK\a- s 


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JIJL 7 190b 

y. Gopyriitnt tnir* 

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/20 / $1 

COPY b. 


Copyright, 1905 
BY 

G. J. FIEBEGER 


rORF.RT DRUMMOND, FRINTF.R, NEW YORE 







PREFACE. 


This text-book is designed primarily for the cadets of the 
U. S. Military Academy, who are being fitted for a profession in 
which the principles of civil engineering are of daily application. 

In time of peace the officer in an isolated station finds himself 
called upon to act as engineer and constructor of buildings, roads, 
and bridges. If not the engineer charged with the construction of 
water-works and sewerage systems, he finds himself charged with 
their maintenance and repair. In time of war, a knowledge of the 
construction of buildings and bridges enables him to effect their 
destruction without loss of time or the aid of skilled workmen. 
The entire subject of military engineering, including fortification, 
sapping, mining, pontoneering, etc., is simply the application of 
the principles of civil engineering and tactics to military problems. 

For the above reasons the Academic Board (Faculty) of the 
Military Academy has, since its origin, provided for a short course 
in civil engineering. The aim of the course is to teach the cadets 
the general principles governing the construction of engineering 
structures, without any attempt to make them specialists. The 
last must be effected in the postgraduate schools. In this course it 
has been necessary to avoid elaborate details and also to omit some 
of the branches of what is now classed as civil engineering, in con¬ 
tradistinction to electrical, mining, and mechanical engineering. 

To fix the principles, problems are given under each head; 

in solving these problems use has been made of the Cambria hand- 

• • • 
m 



IV 


PREFACE. 


book for 1904 , to which references are made. Similar data may, 
however, be found in other handbooks. Additional problems, 
collected in a separate book, are employed in section-room work. 

Descriptions and illustrations from current engineering pub¬ 
lications are constantly used to supplement the text and explain 
the practice of engineering. 

A short list of standard publications is appended to some of 
the chapters. This has been inserted for the use of the officer who 
finds himself in a position in which more exhaustive information 
on a subject is desirable. It is proposed to change these lists from 
time to time. 

I desire to acknowledge my indebtedness to the authors whose 
works have been consulted, and to my assistants Captains James 
P. Jervey, Frederick W. Altstaetter, James A. Woodruff, Horton 
W. Stickle, and Lewis H. Rand, Corps of Engineers, U. S. Army, 
for their assistance and suggestions. 

G. J. Fiebeger. 

West Point, N. Y., May 9, 1905. 



CONTENTS. 


CHAPTER I. 

CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 

PAGE 

General conditions of equilibrium.—Operations in structural design.— 
Non-concurrent and non-parallel forces.—Classification of stresses. 

—General laws of elasticity.—Limit of tensile elasticity.—Coeffi¬ 
cient, or modulus, of tensile elasticity.—Stress-strain curve.— 
Modulus of tenacity.—Compressive stress.—Limit and coefficient of 
longitudinal elasticity.—Modulus of crushing.—Allowable or safe 
unit stress.—Work performed by a longitudinal force.—Relation 
between work, strains, and stresses of suddenly and gradually applied 
forces.—Resilience.—Stress due to impact force.—Elongation of a 
rod of uniform cross-section.—Determination of form of a rod of 
uniform strength.—Determination of thickness of thin cylinders and 
spheres to resist internal pressure.—Longitudinal stresses due to 
changes of temperature. .. i 


CHAPTER II. 

SHEARING AND TORSIONAL STRESS. 

Simple shearing force.—Modulus of shearing.—Safe unit stress.—Shear 
in a rod or column.—Shearing stress in plane perpendicular to plane 
of applied force. — Torsional moment.—Torsional stress. — L T nit 
stress in torsion.—Limit of torsional elasticity —Modulus of torsion. 

Safe unit stress.—Elastic torsional resilience.—Power transmitted by 
shafts. 32 


CHAPTER III. 

FLEXURE OR BENDING. 

Bending moment.—Vertical or transverse shear.—Theorv of bending 

O J o 

or flexure.—Emit stress in flexure.—Modulus of flexure.—Factor of 


v 




VI 


CONTENTS. 


PAGE 

safety.—Section modulus.—Beams of uniform cross-section.— 
Beams of uniform strength.—Bending moment in cantilevers and 
beams resting on end supports.—Dimensions of beams of uniform 
cross-section for simple loading.—Strongest beam cut from a log. 45 


CHAPTER IV. 

BEAMS FIXED AT THE ENDS AND CURVE OF MEAN FIBER. 

Condition fixing end of a beam.—Equation of mean fiber.—Equation 
of mean fiber of a cantilever, of a beam resting on end supports, of 
a beam fixed horizontally at both ends, of a beam fixed horizontally 
at one end.—Table of maximum values of bending moments, unit 
stresses, and deflections in beams.—Stiffness of beams.—Strength 
of beams.—Resilience.—Impact. 60 


CHAPTER V. 

CONTINUOUS BEAMS, HORIZONTAL SHEAR, COMBINED 
STRESSES, AND ECCENTRIC LOADING. 

Continuous beams.—Theorem of three moments.—Beam resting on 
three supports.—Beam resting on four supports.—Beam resting on 
five supports.—Horizontal shear in beams of uniform cross-section. 

—Combined stresses.—Tension and flexure.—Compression and 
flexure.—Flexure and shear.—Longitudinal stress and torsion.— 
Distribution of stress under eccentric loading.—Distribution of 
stress in masonry walls eccentrically loaded. 90 

CHAPTER VI. 

MOVING OR LIVE LOADS ON A BEAM. 

Bending moment due to—single concentrated load, two concentrated 
loads uniform load longer than beam—uniform load shorter than 

beam.—Shear due to—single concentrated load, uniform load 
longer than beam. _ _ 


CHAPTER VII. 

« 

COLUMNS. 

Euler’s formula.—Gordon’s formula.—Rankine’s formula.—Working 
formulas.—Factors of safety.—Eccentric loading.—Built-up col¬ 
umns.—Column design... 







CONTENTS. 


Vll 


CHAPTER YIII. 

RIVETS AND PINS. 

PAGE 

Rivets.—Riveted joints.—Riveted tension-joints.—Pitch and size of 
rivets.—Compression-joints.—Width of main plates.—Thickness of 
cover-plates.—Pins and pin-joints.—Eye-bars.—Design of pins.— 
Maximum shearing and bending moments.—Design for shear, 
flexure, and bearing.—Bearing value of reinforced web. 158 

CHAPTER IX. 

SOLID-BUILT BEAMS, I BEAMS, AND PLATE GIRDERS. 

Solid built beams.—Rolled I beams. Plate girder.—Flanges.—Web.— 
Web-rivets.—Flange-rivets.—Stiffeners.—Box girders.—Design of 
plate girder.—Maximum stresses.—True value of the effective 
depth.—Design of web.—Rivets connecting the flange-angles and 
web.—Flange-plates.—Girder considered as a solid beam. 174 

CHAPTER X. 

DETERMINATION OF STRESSES BY ANALYTIC METHODS 
IN SIMPLE TRUSSES RESTING ON END SUPPORTS AND 
CARRYING UNIFORM DEAD LOADS. 

Methods of determining the stresses in the members of a truss.—Ana- 
Ivtic method of determining reactions.—Analytic method of con- 
current forces.—Stresses in trusses with parallel chords.—Stresses 
in trusses without parallel chords.—The analysis of moments.— 
Method of moments in roof-trusses.—Method of sections. 191 

CHAPTER XL 

EFFECT OF MOVING LOADS UPON TRUSSES WITH 

PARALLEL CHORDS. 

Concentrated load alone.—Dead and concentrated loads combined.— • 
Moving load of uniform weight per unit of length.—Counterbraces.— 
Queen-post truss.—Truss loading. 213 

CHAPTER XII. 

GRAPHIC METHOD OF DETERMINING THE STRESSES 

IN A TRUSS. 

The simple truss problem.—Representation of forces.—Concurrent 
system—Graphic nomenclature.—Other uses of the force polygon. 

—Non-concurrent forces.—Uses of the equilibrium polygon.— 
Truss acted upon by vertical weights alone.—Truss subjected to 
vertical loads and to wind pressure. 225 








X 


CONTENTS. 


PAGE 

cotta.—Tiles.—Glazed and enameled bricks.—Vitrified sewer-pipe 
and blocks. — Common lime. — Manufacture. — Preservation. — 
Tests. — Hydraulic lime. — Cement. — Manufacture. — Preserva¬ 
tion. — Soundness. — Fineness. — Tensile strength. — Specific 
gravity. — Activity. — Short-time tests. — Sand. — Gravel. — Broken 
stone.—Common lime mortar.—Cement mortar.—Uses.—Cement 
concrete.—Use.—Strength.—Artificial cement stones.—Asphalt and 
coal-tar. 361 


CHAPTER XXI. 

MASONRY. 

Rankine’s rules.—Character, size, and shape.—Accuracy of dressing.— 
Bond.—Simple walls.—Weight and allowable compressive stress.— 
Retaining- and reservoir-walls. — Arches. — Centers. — Construc¬ 
tion.—Use.—Pointing masonry.—Settling of masonry.—Effect of 
temperature.—Brick masonry.—Bond.—Walls.—Weight and allow¬ 
able pressure.—Retaining- and reservoir-walls.—Arches.—Strength. 
—Concrete masonry.—Walls.—Allowable pressures.—Retaining- 
and reservoir-walls.—Arches.—Reinforced concrete.—Columns.— 
Rectangular beams.—With reinforcement of tensile fibers only.— 
Area of reinforcement.—With double reinforcement.—Shear.—Ad¬ 
hesion.—Methods of reinfo cing.—Uses.—Arches. 


CHAPTER XXII. 

FOUNDATIONS. 

General principles.—Classes and bearing powers of soil.—Testing soils._ 

In firm soils.—Soft soils.—Piles.—Pile-driving apparatus.—Com¬ 
mon pile-driver.—Steam-hammer—The water-jet.—The capstan. 
Load on piles.—Theoretical formulas.—Method of constructing 
common pile foundations.—Random rock.—Cribs.—Concrete.— 
Piles.—The Cushing cylindrical pier.—The common caisson.— 
The well or open dredging process.—Methods of sinking.—Dredg- 
ing apparatus. Poughkeepsie bridge.—The Hawksbury bridge — 
The diving-suit.—Common dams.—Coffer-dam.—Modifications of 
the coffer-dam.—Pneumatic caisson.—Construction.—Sinking the 
caisson.—The caissons of the New York suspension bridges.—The 
Forth bridge. 


414 





CONTENTS. 


XI 


CHAPTER XXIII. 

BRIDGES. 

PAGE 

Definitions. — Wooden trestles. — Steel trestles and towers. — Cribs. 
Masonry piers.—Approaches and abutments.—Floor system.— 
Metal floors.—Camber.—Floor-carriers.—Plate girder.—Truss.— 
Cantilever truss.—Howe truss.—Arch bridges.—Masonry arches.— 
Form of intrados.—Spandrel.—Capping.—Luxemburg bridge.— 
Arches of reinforced concrete.—Steel arches.—Suspension bridges. 

—Main cables. — Towers. — Anchorages. — Suspenders. — Oscilla- 
lations.—Floor.—East River bridges.—Loads on bridges.—Impact. 

— Wind stresses. — Wind-bracing. — Sway-bracing. — Temperature 
stresses.—Bridge erection.—Drawbridges. 442 

CHAPTER XXIV. 

TRUSSED ROOFS AND FLOORS. 

Roofs.—Definitions.—Methods of construction.—Dead and live loads.— 
Wind pressure.-Floors.—Different forms.—Methods of con¬ 
struction.—Dead and live loads. 472 

CHAPTER XXV. 

HIGHWAYS. 

Definitions.—Profile.—Tractive power of a horse.—Weight of load — 
Tractive resistance of vehicles.—Maximum and minimum grades.— 
Cross-section. — Roads. — Streets. — Location. — Excavations.— 
Embankments.—Culverts.—Side ditches.—Subsoil drains.—Water- 
breaks.—Catch-basins.—Earth roads.—Gravel roads.—Macadam. 

—Binder.—Stone.—Rolling. — Telford-macadam. — Comparison of 
Telford and Macadam methods.—Wearing surfaces.—Foundations. 

—Wood pavements.—Stone-block pavement.—Brick pavements.— 
Asphalt pavements.—Comparison of pavements. 481 

CHAPTER XXVI. 

WATER-SUPPLY. 

Character of supply.—Amount required.—Measuring the supply.—Varia¬ 
tion in flow of streams.—Variation in flow of springs and wells.— 
Receiving-reservoirs.—Dams.—Spillway.—Masonry and rock-fill 
dams. — Intake. — Conduits. — Pumping-machinery. — Selection of 
sources of supply.—Settling-basins.—Filters.—Slow filters.—Rapid 
filters.—Disti*ibuting-reservoir.—Water-mains.—Meters. 510 







SYMBOLS EMPLOYED IN MECHANICS OF ENGINEERING. 


y 4 =area of cross-section. 
b = breadth. 

7 = depth. 

E, E lt ^^coefficients or moduli of longitudinal elasticity. 
Et= coefficient or modulus of lateral elasticity. 

F = intensity of a force. 

/' = coefficient of expansion. 
g = acceleration due to force of gravity. 

Ha = horizontal shear in a beam. 

1 = moment of inertia of the cross-section of a beam 
about its neutral axis. 

7 p--= moment of inertia of the cross-section of a shaft 
about the axis of the shaft. 


— = section modulus of a beam. 

y 

l, L = length. 

M= bending moment. 

M m — maximum bending moment. 

Mt= torsional moment. 
r, R= radius of circle or radius of gyration. 

R i, R2 = reactions at supports of a beam. 

s=unit stress of flexure in surface fiber of beam. 
s'"— “ “ “ “ at a unit distance from the 


neutral axis. 

s c =unit stress of compression. 

s e = u “ “ elongation or tension. 

j«= “ “ “ shear. 

st = 11 “ “ torsion in surface fiber of a shaft. 

st'" = “ “ “ “ at a unit distance from the 

axis. 

s", s c ", s e ", s 8 ", etc. =safe values of the unit stresses, 
s', Sc', Se', s 8 ', etc. ^ultimate or breaking values of the unit stresses. 

Sm = maximum value of s. 

V s = vertical shear in a beam. 

( V s ) m = maximum vertical shear. 

W —• weight or load. 

y' = distance of surface fiber of abeam from the 
neutral axis. 

y m = maximum deflection of a beam. 
a t & r, and 0= angles. 

<? = sign of differentiation. 



= sign of integration. 


X = unit elongation or shortening, 
inch-pounds = work. 

(inch-pounds) = moment of a force. 

xiv 


CIVIL ENGINEERING. 


CHAPTER I. 

CLASSIFICATION OF STRESSES AND LONGITUDINAL 

STRESS. 

An engineering structure, as a building, bridge, dam, etc., 
is one designed on the hypothesis that its different elementary 
parts are to be in a state of rest. It differs in this respect from 
a machine, which is designed to admit relative motion between 
some of its parts. 

Rest in a body is possible only when the system of forces 
acting on it is in equilibrium. In order that the structure as 
a whole shall be in a state of rest, the system of forces acting on 
the entire structure must be in equilibrium. In order that any 
elementary part of the structure shall be at rest, the system of 
forces acting directly on it, or transmitted to it through the other 
parts, must be in equilibrium. 

The general conditions of equilibrium for any system of 
forces are: 

1. The algebraic sum oj the components oj all the forces in any 
direction must be equal to zero. If this condition is fulfilled, the 
center of the body acted on can have no linear motion or motion 
of translation. 

2. The algebraic sum of the moments of the forces about any 
■axis must be equal to zero. If this condition is fulfilled, the body 
can have no motion of rotation about any axis. 



2 


CIVIL ENGINEERING. 


In most engineering structures, it is assumed that the system 
of forces acting on each elementary part, and sometimes on the 
entire structure, is a coplanar system; that is, the action lines 
of all the forces of the system lie in the same plane. If the plane, 
is vertical, the coplanar forces are usually resolved into vertical 
and horizontal components. The general conditions of equilib¬ 
rium for a coplanar system of forces in a vertical plane are: 

1. The sum oj the vertical components oj the forces must be 
equal to zero. 

2. The sum of the horizontal components of the forces must be 
equal to zero. 

3. The sum of the moments of the forces, about any axis per¬ 
pendicular to the plane of the forces, must be equal to zero. 

Operations in Structural Design. —The process of structural 
design may be divided into three operations: 

1. The determination of the intensity and action line of 
every force that acts on the structure as a whole, or on any one 
of its elementary parts. 

2. The determination of the behavior of the material of which 
the structure is made, under the application of force. 

3. The designing of each elementary part, after the system 
of forces acting on it, and the behavior of its material under the 
application of force, are known. 

First Operation. —The problems that arise under the first 
operation are usually those of coplanar forces and may be solved 
by the aid of the three equations of condition for equilibrium 
in a coplanar system. These are: 

A sin a + B sin /? + C sin y+D sin o + etc.=o, . . . (1) 

A cos a + B cos/? + Ccos y+D cos <5 + etc.=o, ... (2) 

AV sin a + Bl" sin [ 3 +Cl'" sin y+Dl™ sin £ + etc.'=o, (3) 

in which A, B,C, D, etc. = the intensities of the different forces 

of the system; 

a > A T> etc. = the angles which the action lines of 

the forces make with the horizontal; 


CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS . 3 


V, / IV , etc. = the distances from the center of 

moments to the action lines of the 
different forces, measured on a 
horizontal line through the center 
of moments; 

sin a, etc. = the lever-arms of the forces with 
respect to the center of moments. 

In the above equations there are three quantities dependent 
on each force. Thus A is the intensity of the force A, the angle 
a fixes the direction of its action line in the plane, and the dis¬ 
tance l f fixes the position of the action line in the plane or a point 
oj application of the force A. 

Since each intensity enters each of the three equations, we 
may by combination determine three unknown intensities if all 
the other quantities in the equations are known and the equa¬ 
tions are independent. 

Similarly we may determine three unknown directions or 
angles, if all the other quantities are given and the equations 
are independent. The combination of the equations is, however, 
difficult. 

Since the quantities V", etc., enter but a single equa¬ 

tion, (3), we can determine but a single unknown point of appli¬ 
cation when all the other quantities in this equation are given. 

In combining the equations, components acting upwards or 
to the right are considered positive , and those acting in contrary 
directions negative. Moments which are clockwise are considered 
positive , and counter-clockwise moments are considered negative. 

Non-concurrent and Non-parallel Forces.—If the action lines 
of the forces neither intersect at a common point nor are par-, 
allel, equations (1), (2), and (3) are wholly independent and 
we may determine any three unknown quantities in the system 
if all the others are given, provided that only one of the unknowns 
is a point of application. 

Concurrent System.—If the action lines of the forces inter¬ 
sect at a common point, equation (3) is no longer independent 
of the others. This is most readily shown by resolving the forces 
into components parallel and perpendicular to the line joining 


4 


CIVIL ENGINEERING. 


the common point with the center of moments, instead of resolv¬ 
ing them into vertical and horizontal components. The angles, 
a , /?, etc., will then be the angles which the action lines of 
the forces make with this new line, and the distances l", 
etc., will be measured along the line. Since the distances /" 
etc., will all be equal under this hypothesis, equation (3) 
becomes identical with equation (1), if we divide (3) through 
by the common value of l. 

In a concurrent system we can therefore determine but 
two unknown quantities neither of which can be a point of appli¬ 
cation. 

Parallel Forces.—If the action lines of the forces are parallel, 
equations (1) and (2) are no longer independent, since the angles 
a, /?, f, etc., are all equal to each other. By dividing equation (1) 
by the common sine, and equation (2) by the common cosine, 
the two equations become identical. 

In a parallel system we can therefore determine but two 
unknown quantities, of which one only may be a point of appli¬ 
cation. In solving systems of parallel forces, the angles a, /?, 
etc., are usually assumed as 90°, and the distances etc., 

are measured on a line through the center of moments perpen¬ 
dicular to the action lines of the forces. 

For a system of parallel forces the three equations then reduce 
to 

A 4- B -(- C T D -)- etc. = o.. (kz) 

AV + Bl" + Cl nt + DF y + etc.=o. . . . ( 3 d) 

Three Forces.—If three forces are in equilibrium or a body 
is at rest when acted on by three forces only, each force must be 
equal and opposite to the resultant of the other two; hence the 
three forces must form a parallel or a concurrent system. It 
follows that if in any system of forces in equilibrium all are 
known save two, these two must, with the resultant of the 
known forces, form a concurrent or a parallel system. 

Weight.—One of the forces which must usually be considered 
as acting on an engineering structure is its own weight. It is 



CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 5 


therefore necessary to know the weights of the ordinary build¬ 
ing materials. These are approximately as follows: 


Wood. 

Brick masonry. . . 
Concrete masonry 
Stone masonry. .. 

Cast iron. 

Wrought iron. . .. 
Steel. 


Cubic Foot. 

30 to 50 pounds 

100 “ 140 

125 “ 140 “ 

140 “ 160 “ 

.450 

.480 

.490 


Second Operation.—To determine the behavior of materials 
under the action of force, we must first consider the manner in 
which force may be applied. 

If a solid body, as a beam, is firmly fixed, horizontally, in 
a wall (Fig. 1), it may be acted upon— 

1. By a force whose action line coincides with the axis of 
the beam and whose direction may be either away from or 
towards the fixed end of the beam. Such a force is a longitu¬ 
dinal jorce; tensile if it acts away from the fixed end, and com¬ 
pressive if it acts towards the fixed end. Thus in Fig. 1, AB 
is a tensile force and AC is a compressive force. 

A piece designed to resist a tensile force is called a rod or 
tie; a piece designed to resist a compressive force is called a column 
or strut. 

2. By a force whose action line is in the plane of the axis 
of the beam and is perpendicular to the axis. This is a bend¬ 
ing-shearing jorce , usually called 
simply a bending force or a force 
of flexure. Such a force be¬ 
comes a simple shearing force 
when it acts in the plane adjacent 
to the fixed plane of cross-section. 

In Fig. 1 AD is a bending force; 
if the force AD were applied in 
the plane of cross-section adja¬ 
cent to the fixed end of the beam, it would be a simple shearing 

force. 





END 

elfvati^n 


Fig. 1 . 




















6 


CIVIL ENGINEERING. 


A piece designed to resist a bending force is called a beam. 

3. By a force which acts in one of the planes of cross-section 
of the beam, but does not intersect the axis. This is a torsional 
or twisting jorce , and its moment with respect to the axis of the 
beam is called its torsional moment. In Fig. 1 the force EF 
acting in the plane of the end cross-section is a torsional force. 
A torsional couple is a couple formed of two equal torsional forces 
acting in contrary directions. A single torsional force produces 
bending as well as torsion; a torsional couple produces torsion 
only. If in Fig. 1 a parallel force equal to EF acts upwards 
in the same cross-section, the two together will form a torsional 
couple. 

A piece designed to resist a torsional force is called a shaft. 

Any system of forces acting on an elementary piece of a 
structure may be resolved into these elementary forces. This 
resolution is usually necessary before the effect of the system 
can be determined. 

If a force intersects the axis of the beam in Fig. 1, but is oblique 
to that axis, it may be resolved into a bending force and a ten¬ 
sile or compressive force. If a force is perpendicular to the 
axis but does not intersect it, it may be resolved into a bending 
force and a torsional couple. Thus the force EF in Fig. 1 may 
be resolved into or replaced by an equal bending force acting 
downwards through G, the center of gravity of the cross-section, 
and a couple consisting of the original force EF and an equal 
force acting upwards through G. A force which is oblique to 
the axis which it does not intersect may be resolved into a bend¬ 
ing force, a torsional couple, and a longitudinal force. 

Classification of Stresses.—The effect of a force on a body 
in a state of rest is to move its molecules with respect to each 
other. This disturbs the equilibrium of the molecular forces, 
and results in the establishment of new' conditions of equilibrium 
in which the molecular forces developed by the change in the 
relative positions of the molecules are in equilibrium with the 
force which has caused the change. 

To distinguish the force which is applied to the body from 
the molecular forces which hold the applied force in equilibrium, 


CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 7 


the former is often called an applied or extraneous force, and 
the latter are called stresses or internal stresses .* 

1. Under the action of the force AB or the force AC (Fig. i) 
every molecule in the beam will move parallel to the axis of the 
beam. The resulting stress is called a longitudinal stress; ten¬ 
sile if the molecules are pulled apart, and compressive if pushed 
closer together. In Fig. i the force AB will develop tensile 
stress, and the force ^4C compressive stress. 

2 . Under the action of the force AD or of the force EF 
(Fig. i), every molecule of the beam will move in its plane of 
cross-section. The resulting stress is called a lateral stress; 
shearing if the molecule moves in a right line, and torsional if 
the molecule moves in a circular path. In Fig. i the force AD 
will develop shearing stress, and the force EF torsional stress. 

The space passed over by a molecule in longitudinal stress 
is called its strain* and that passed over by a molecule in lateral 
stress is called its distortion* 

The path of any molecule of the beam under the action of 
any applied force may be resolved into its component strains 
and distortions. 

The determination of the behavior of material under each 
of the elementary forces is usually effected in a testing labora¬ 
tory. For each kind of force it is necessary to determine the 
law governing the distribution oj the stress developed by it over 
the plane oj cross-section. This law is general and applies to all 
materials. It is also necessary to determine for each kind oj 
jorce the elastic properties developed by it in each kind of material , 
and the ultimate and allowable strength oj each kind oj material. 

These results are usually tabulated for use in designing the ele- 

% 

mentary parts of a structure. 

Tensile Stress. —If a steel rod is suspended from a ceiling 
and to the lower end of its axis is attached a weight, the rod 
will be subjected to a tensile force. Laboratory experiments 
show— 


* By some engineers the term stress is applied to any force which acts 
on a body in a state of rest; the term strain to an internal stress; and the 
term distortion to a strain. 




8 


CIVIL ENGINEERING. 


1. That all the fibers between the ceiling and weight are 
elongated. 

2. Every cross-section normal to the axis of the rod before 
the weight is applied remains normal to the axis after the weight 
is applied. 

3. That these effects are observed whatever be the form of 
cross-section of the rod. 

It follows, therefore, since the material is homogeneous and 
all fibers have the same resistance, that the stress on the cross- 
section must be uniform and hence on each unit of area the 
same. 

Let F = total tensile force in pounds applied to, or the total 
tensile stress developed by it in, any cross-section; 
A = area of cross-section considered, in square inches; 
s e = tensile force or tensile stress in pounds per square 
inch, called the unit tensile jorce or unit tensile 
stress. Then 


F_ 

Se ~ A .* * • * ( 4 ) 

In the English system the unit tensile force and unit tensile stress 
are generally expressed in pounds per square inch. 

The law governing the distribution of the stress over the 
area of cross-section, and expressed in equation (4), is general 
for all materials under tensile stress, whatever be the form of 
cross-section, providing the force is applied along the axis of 
the rod. Unless great accuracy is desired, it is also assumed 
to be true for non-axial forces. 

General Laws of Elasticity.—The general laws governing the 
elasticity developed in rods and columns by longitudinal forces 
as derived from experiments are: 

1. All bodies are elastic. 

2. Within a limited field all bodies are perfectly elastic. 

3. Within the field of perfect elasticity , the strain varies directly 
with the intensity of the applied force. 

4. Beyond the field of perfect elasticity , the strain increases 


CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 9 

more rapidly than the applied jorce until rupture takes 
place. 

Limit of Tensile Elasticity.—Conceive a rod of structural 
steel one square inch in cross-section, hung from the ceiling of 
a room, and supporting at its lower extremity a scale-pan for 
weights. If this pan is loaded with a gradually increasing load, 
the bar will stretch proportionally to its load until the aggre¬ 
gate of the suspended weights reaches about 35,000 pounds. 
During the loading, if the weights are, at any time, temporarily 
removed, the bar will at once return to its original dimensions. 
If the weights are increased above 35,000 pounds, the ratio 
of the elongation to the increase in the weights will no longer 
be constant, nor will the bar return to its original dimensions 
when the weights are temporarily removed. This indicates 
that a unit stress of 35,000 pounds limits the field of perfect 
elasticity of structural steel. It is therefore called the limit 
oj tensile elasticity of that metal. 

If, when the weights exceed 35,000 pounds, they are tem¬ 
porarily removed, it will be observed that the bar has received 
a slight permanent elongation; this elongation is called a per¬ 
manent set. 

The limit oj tensile elasticity oj any material is the greatest 
unit tensile jorce which may be applied to, or the greatest unit 
tensile stress which may be developed in, a rod oj the material 
without producing a permanent set. 

Coefficient, or Modulus, of Tensile Elasticity.—Let 

F = tensile force in pounds applied to, or the tensile stress 
developed in, any area of cross-section of a rod; 

yl=area of cross-section in square inches; 

L = original length of the rod in inches; 

/ = elongation of the rod in inches. 

17 

Then _ j s t h e un it jorce applied to, or unit stress developed 
A 

in, the area of cross-section, and -jr is the unit elongation or the 

elongation per unit of length; it is usually represented by X. 

By the third general law of elasticity, the unit elongation 


10 


CIVIL ENGINEERING. 


must, within the field of perfect elasticity, vary directly with 
the unit force or unit stress; hence 


F 



( 5 ) 


L 


in which E is a constant. 

F l 

If as before we make ^ = ^> equation (5) becomes 



( 6 ) 


Since the second form of equation (6) is of the same form 
as x = ay , the equation of a right line, in which a is a constant 
coefficient, E is called the coefficient oj elasticity. From the 
third form of equation (6) it is evident that for any unit 
stress or force, within the field of perfect elasticity, we may 
determine the corresponding elongation for any material whose 
value of E is known; it is therefore also called the modulus oj 
elasticity. 

From the first form of equation (6) it is seen that since X 
is merely an abstract fraction, E must be expressed in the same 
units as s e , or in pounds per square inch. 

The coefficient, or modulus, oj tensile elasticity oj any material 
is the jorce or stress obtained by dividing any unit tensile jorce or 
unit stress, not exceeding its limit oj tensile elasticity, by the corre¬ 
sponding unit elongation. 

Stress-strain Curve.—If from any origin of rectangular 
coordinates we lay off on the axis of ordinates the values of 
the successive tensile forces applied to a steel rod, and on the 
axis of abscissas the corresponding elongations, the resulting 
curve will be a stress-strain curve. 

If the curve is plotted automatically by a testing-machine, 
it will be of the general form shown in Fig. 2. Within the field 
of perfect elasticity it will be a right line, since its equation is 




CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. H 


F = al, in which a is constant. Beyond the field of perfect 
elasticity a becomes variable and the equation is no longer that 
of a right line. 

The ordinates of the curve will increase until rupture takes 
place. The point of rupture of a ductile metal, like structural 
steel, is marked by the running of the metal and a decrease 
of cross-section at the point of rupture. This is indicated on 
the stress-strain curve plotted automatically by a testing-machine, 
by a dropping of the curve towards the axis of elongations, and 
by the final tearing apart of the decreased section by a force less 
than the maximum applied force. If the material is not ductile, 
the points of rupture and final separation are coincident. 

If the plotted ordinates are the successive applied forces 

F 

divided by the original area of cross-section, or -j, and the 
original length of the rod is assumed as ioo units, and the plotted 
abscissas are the resulting curve is that shown in Fig. 2. The 



Fig. 2. 

ordinates are unit forces or stresses, from the origin to the point 
of separation, and the abscissas are unit elongations or strains. 

The point B of the curve where the elongations suddenly 
increase is called the yield-point . The ordinate of the point 
just below the yield-point, where the straight portion of the stress- 
strain curve ceases, is the limit of elasticity. 
















































































































12 


CIVIL ENGINEERING. 


Modulus of Tenacity.—If the load on the one-inch rod of 
structural steel, page 9, is increased until it reaches about 70,000 
pounds, the rod will tear apart and the weight will drop to the 
floor. This indicates that a unit stress of 70,000 pounds measures 
the tensile strength of this metal; it is therefore called the modulus 
oj tenacity , or the ultimate strength 0j the material in tension. It 
will be represented by sj. 

The modulus oj tenacity oj any material is the unit tensile 
force applied to , or the unit tensile stress developed in , a rod oj 
the material at the moment oj rupture. In the stress : strain curve 
shown in Fig. 2 it is the ordinate of the highest point of the curve. 

Compressive Stress.—If a solid steel column whose cross- 
section is a circle or a regular polygon, and whose length does 
not exceed ten times its least dimension of cross-section, is placed 
in a vertical position and loaded, it will be observed that— 

1. All the fibers between the support and load are shortened. 

2. Every cross-section normal to the axis of the column before 
the weight is applied remains normal to the axis after the weight 
is applied. 

3. That these effects are observed whatever be the regular 
form of the cross-section. 

It follows, therefore, that the stress on each cross-section 
must be uniform and hence on each unit of area the same. 

Let F = total applied compressive force, or total compressive 
stress in pounds, on any cross-section of a short 
column; 

A = area of cross-section in square inches; 
s c = compressive force or stress in pounds per square 
inch. Then 

F 



for the unit stress of compression. The law of distribution of 
compressive stress in short columns as expressed in equation 
(7) applies to all materials. 

Limit of Compressive Elasticity. —The limit oj compressive 
elasticity may be determined in a manner similar to that of ten- 



CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 13 


sile elasticity and is the greatest unit compressive jorce which 
may be applied to , or the greatest unit compressive stress which 
may be developed in , a short column oj the material without pro¬ 
ducing a permanent set. 

Coefficient, or Modulus, of Compressive Elasticity.—Let 

F = the compressive force applied to, or the compressive stress 
developed in, the cross-section of a short column, ex¬ 
pressed in pounds; 

A= area of cross-section in square inches; 

L = original length of the column in inches; 

/ = the amount of shortening in inches. 

Then, as under the action of a tensile force, we shall have 



E = 



s c = EX, 


and A=4?, 
E 



in which E is the coefficient or modulus of compressive elasticity. 

This coefficient may be defined as the jorce or stress obtained 
by dividing any unit compressive jorce or stress , not exceeding the 
limit oj elasticity oj the material considered , by the corresponding 
unit shortening oj the column to which it is applied. 

Stress-strain Curve.—A stress-strain curve may be con¬ 
structed for compression in the same manner as that for ten¬ 
sion. When both curves for the same material are plotted on 
the same diagram, it is usual to lay off compressive forces from 
the origin downwards and shortenings to the left. The tensile 
curve is then in the first, and the compressive curve in the third, 
quadrant. 

Limit and Coefficient of Longitudinal Elasticity.—It is found 
by experiment that the limits and coefficients of elasticity for 
tension and compression of ordinary structural materials differ 
so little from each other that common values of these constants 
may be employed in all practical problems. These common 
values for any material are called its limit and its coefficient of 
longitudinal elasticity. 


14 


CIVIL ENGINEERING. 


Modulus of Crushing.—The modulus oj crushing or ultimate 
strength in compression is determined in a manner similar to 
that employed in determining the modulus of tenacity and is the 
unit stress which is developed in a column oj material at the 
moment oj rupture , or the unit jorce which produces rupture. The 
column must be so short that the material is crushed without 
bending its fibers. If of steel, a cylindrical column will give 
way by crushing if its diameter is at least one-tenth its length; 
a wooden column must have a diameter at least one-fifth its 
length. The modulus of crushing will be represented by s c '. 

Allowable or Safe Unit Stress .—The allowable or saje unit 
tensile stress is the greatest unit stress to which it is deemed 
advisable to subject a material. For all materials it must be 
less than the modulus of tenacity, and for ductile metals it is 
usually less than the limit of longitudinal elasticity. In com¬ 
pression the allowable or safe unit stress must be less than the 
modulus of crushing, and for metals it is usually less than the 
limit of longitudinal elasticity. The safe unit stress in tension 
or elongation will be represented by s e " and in compression by s c ". 

The following table gives general values of the constants 
above described, in pounds per square inch, for ordinary build¬ 
ing materials. They vary, however, considerably for different 
varieties of the same material. 



Wood. 

Stone. 

Cast Iron. 

Wrought 

Iron. 

Steel. 

Limit of longitudinal 
elasticity. 

3,°oo 


6,000 

2 5,000 

35,ooo 

Coefficient of longitudi- 




nal elasticity. 

1,500,000 

1 , 000,000 

15,000,000 

2 7 , 000,000 

30,000,000 

Modulus of tenacity. . . . 

10,000 

500 

20,000 

50,000 

75,000 

Safe unit stress in ten- 




sion. 

1,000 

50 

3,000 

10,000 

12,000 

Modulus of crushing. . . . 

6,000 

5,000 

80,000 

50,000 

50,000 

Safe unit stress in com- 





pression. 

800 

600 

10,000 

T 0,000 

1 2,000 


Third Operation.—The third operation in structural design 
consists in, determining the dimensions of the elementary pieces 


























CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 15 


of a structure when the forces which it is required to resist 
are fully determined, and the constants of the material are 
known. 

If the piece is a rod which must resist a tensile force, or a 
short column which must resist a compressive force, the opera¬ 
tion is usually that of finding the minimum area of cross-section 
which will safely resist the applied force. 

Let F = the applied tensile or compressive force-, in pounds; 

A = the required area of cross-section in square inches; 
5/' = safe unit stress in tension; 
s c " = safe unit stress in compression. Then 

F F 

-4=777 or 777.. (9) 


From these equations we may determine any one of the 
quantities when all the others are given. 

It is also sometimes necessary to know the amount of elon¬ 
gation or of shortening which will occur in a rod or column under 
the action of a force which will produce a unit stress not exceed¬ 
ing the limit of elasticity. 

Let F = applied force in pounds; 

A = area of cross-section in square inches; 

L = length of rod or pillar in inches; 

/ = elongation or shortening in inches; 

E — coefficient of elasticity. Then 


From this equation we may determine any one of the five 

quantities when the others are known. 

The values of the constants to be used in solving problems 
are those given in the tables. The weight of the piece itself 
will be omitted unless specifically stated otherwise. 






CIVIL ENGINEERING. 


16 


PROBLEMS. 

1. Determine the diameter of a standard* circular wrought- 
iron rod which will safely support a weight of 100,000 pounds. 

Ans. 3! inches. 

2. What load may be safely borne by a standard * wrought- 

iron rod f- inch in diameter? Ans. 6013 pounds. 

3. How much will a hundred-foot steel tape, 7 inch wide and 

so inch thick, stretch under a pull of 50 pounds ? . 4 ns. 0.2 inch. 

4. Find the coefficient of elasticity of wrought iron of which 

a rod 1 inch in diameter f and 11 feet long will stretch l inch under 
a tensile force of 20,000 pounds. Ans. 26,890,756. 


Work Performed by a Longitudinal Force.—The force applied 
to elongate a rod or compress a column may be either a gradu¬ 
ally or a suddenly applied force. A gradually applied force is 
one whose intensity gradually increases from zero to a maximum; 
a suddenly applied force is one whose intensity is constant. 

Work of Gradually Applied Force.—The expression for the 
work done by a force of variable intensity, as a gradually applied 
force, is 


work = J Foly 


(10) 


in which F = variable intensity of the force; 

/ = path of the force. 

If F is a longitudinal force, and l is its strain, equation (10) 
gives the work of elongation or compression. If a stress-strain 
curve is constructed from corresponding values of F and /, the 

expression j Fol will measure the area included between the 

curve itself, the axis of elongations or abscissas, and the 
ordinate of the point corresponding to the maximum strain 
considered. Thus in Fig. 2 the area included between the 
stress-strain curve, the axis of elongations, and the ordinate 
of the highest point of the curve measures the work done in 
rupturing a rod of steel of a unit cross-section. The work neces- 


* Cambria Handbook, pp. 383-389. 
f Ibid ., p. 407, areas of circles. 







CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 17 


sary to tear the rod apart is measured by the area included between 
the curve, the axis of abscissas, and the ordinate of the point C. 
If L in Fig. 2 is assumed as 100 inches, each small rectangle will 
represent 5 000 inch-pounds. It will be observed that the work 
required to stretch the rod to its elastic limit is but a small frac¬ 
tion of the work of rupture, although the modulus of rupture is 
only double the elastic limit. 

Suddenly Applied Force.—The expression for the work done 
by a force of constant intensity, as a suddenly applied force, is 

work = F f ol, .(11) 


in which F = intensity of force; 

/ = path of force. 

If the force is a longitudinal one, equation (n) gives the 
total work of elongation or compression. 

In Fig. 2 the work performed by a suddenly applied force 
is represented by the rectangle whose height is the intensity of 
the unit force and whose length is the temporary elongation 
produced by it. 

Relation between Work, Strains, and Stresses of Suddenly 
and Gradually Applied Forces.—If a pound weight is placed in 
a scale-pan of a spring balance and the pan finally comes to rest 
one inch below its original position, the final work performed will 
be one inch-pound; the temporary work will be one pound multi¬ 
plied by the total distance the pan temporarily descends. If a 
pound of sand is poured into the same scale-pan grain by grain, 
the final and temporary work will be the same, and each equal to 
one inch-pound. Therefore the final work of a suddenly applied 
and a gradually applied force of the same intensity will be equal; 
but the temporary work of a suddenly applied force will be greater 
than that of a gradually applied one of equal intensity. 

From equation (10) we may determine the work performed 
by a gradually applied force by substituting for F its value in 
terms of l from equation (5), and integrating between proper 
limits; or 



i8 


CIVIL ENGINEERING. 


temporary work 1 pv EA pi' 
final work \ ^ i Fol = irJo ldl 


EAP F’l' 
2L 2 * 


(12) 


F'V 

— is the temporary and also the final work performed by the 

gradually applied force whose maximum intensity is F', in pro¬ 
ducing a strain whose maximum value is l'. 

From equation (n) we may determine the temporary work per¬ 
formed by an equal suddenly applied force F', in producing the 
temporary strain If by making F = F' and / = /'; thus, 


temporary work — F'V .(13) 


The final work performed by the suddenly applied force 
F' must be equal to that of the gradually applied force, or 


final work = 


F'l' 

2 


. . (14) 


From equations (12) and (13) it follows that— 

1. The temporary work performed by a suddenly applied longi¬ 
tudinal force , within the field oj perfect elasticity , is twice that 
performed by a gradually applied load of the same maximum 
intensity. 

If the work of a gradually applied force is equal to that of a 
suddenly applied one, this can result only by doubling either the 
intensity of the force or its strain. From equations (12) and (13) 
we have 


temporary work = 


( 2 F’)V 

2 




Hence— 

2. A suddenly applied load will , within the field of perfect 
elasticity, produce a temporary strain as great as the permanent 
strain produced by a gradually applied force of double its intensity. 

3. Since within the field of perfect elasticity the unit stress 
varies directly with the unit strain , it follows that if the unit stress 
is not to exceed the limit of longitudinal elasticity or a permanent 







CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 19 


set is not allowable , the unit intensity of a suddenly applied force 
must not exceed one-half the limit of elasticity. This results from 
the definition of the limit of longitudinal elasticity. 

This last conclusion is important in determining the safe 
unit stress on bridge members which must be designed to resist 
suddenly applied loads. 

In equation (12) the work of a gradually applied load is ex¬ 
pressed in terms of /', E, A, and L\ in some problems it is con¬ 
venient to have it in terms of F', E, A, and A; this may be done 
by substituting the value of/' in terms of F' from equation (5), or 


work of gradually applied force 


F' 2 L 
2 EA m 


■ ■ (16) 


PROBLEMS. 

5. The work expended by a gradually applied force in elongating 

a 1-inch square wrought-iron rod 25 feet long is 100 foot-pounds. 
What is the intensity of the force applied and the elongation 
produced? Ans. Force 14,697 pounds. 

Strain 0.167 inch. 

6. A standard circular steel rod is required to support a sud¬ 
denly applied load of 20,000 pounds; what is the minimum diameter 
of such a rod if a permanent set is to be avoided? 

Ans. 1 1 inches. 

Resilience.—The work expended by a longitudinal force in 
straining a rod or column within the field of perfect elasticity 
is equal to the potential energy of elasticity thus stored in it. 
This potential energy of elasticity is called its longitudinal resili¬ 
ence. 

It may be derived from the equation 

EA f 181 

resilience = work = J Fdl =-7-*. . . . (17) 


by integrating between the limits of zero and any value /' less 
than that corresponding to the limit of elasticity; or 


resilience = 


EAV 2 
2L ' 


(18) 






20 


CIVIL ENGINEERING. 


Since 5 C or 



El' 
L ’ 


we have 



E 2 l ' 2 



EAl ' 2 EL 

Multiplying the expression — by and substituting in the 


E 2 !' 2 


resulting equation, s e 2 for ■ ^ 2 ~, we have 


or 


s 2 AL 

resilience = —77-, 
2 E 


resilience ... . . f . s? 

—— = resilience m a unit of volume = 


(20) 

(21) 


The expression ^ becomes the modulus oj longitudinal 

elastic resilience when is increased until it becomes equal to 
the limit of longitudinal elasticity. 

The modulus of longitudinal elastic resilience of any mate¬ 
rial measures the capacity of a rod or column of the material, 
in comparison with similar rods or columns of other materials, 
to store the work expended on it, within the field of perfect elas¬ 
ticity, by a longitudinal force. It is obtained by dividing the 
square of the limit of elasticity of the material by twice its coeffi¬ 
cient of longitudinal elasticity. 


PROBLEMS. 

7. What potential energy in foot-pounds is stored in a steel 
rod 3 inches in diameter and 10 feet long by increasing its unit 
tensile stress from 10,000 to 20,000 pounds? 

Ans. 353.6 foot-pounds. 

8. In increasing the unit tensile stress in a wrought-iron rod 
from 10,000 to 20,000 pounds, 325 foot-pounds of work are expended. 
If the diameter of the rod is 4 inches, what is its length? 

Ans. 4.6 feet. 












CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 21 


Stress Due to Impact Force.—An impact force is the striking 
force of a moving body. If a weight IF falls through a distance h 
before it begins to strain a rod or column, the total amount of 
work performed before it ceases its motion downwards will be 

work = IF(/t + /'),.(22) 

in which IF = weight of IF in pounds; 

h = distance of fall in inches; 

/' = amount of strain in inches. 

If the strain is within the field of perfect elasticity, the work 
of the weight will be equal to the resilience of the rod or column, 
or 

FI' 

work = IF (h-r l ')=—.(23) 

2 

in which F' is the intensity of a gradually applied force which 
produces a permanent strain, which is equal to the temporary 
strain, produced by the falling weight. 

EAl' 

If we substitute for F' its values ^ - and s e A, in which 

5^ = unit tensile stress corresponding to F' and A= area of 
cross-section, we shall have 


F'l' 

W (k+l')= — 


EAl ' 2 s e Al' 
2L ~ 2 * 


• • (24) 


From these equations we can determine any one of the quan¬ 
tities when all the others are known. 


. EAl ' 2 

If we transform the expression —j- 
resilience, we shall have 

s 2 AL 

W(h+n-hw- 


as in the discussion in 


(25) 


o " 

If in this expression we make equal to the modulus of 
longitudinal elastic resilience, /' equal to the strain corresponding 











22 


CIVIL ENGINEERING . 


to the limit of elasticity, and give W, A, and L their values, we 
may determine the greatest height, h , from which W may be 
dropped without producing a permanent set in the rod. 

Ratio of Stress and Strain of Impact and Gradually Applied 
Forces.—From equation (23) we have, making W=F, 


F' F' 2 ( 7 * + /') 
W~F~ l f 



If we represent by / the permanent strain produced by W or F 
acting as a gradually applied force, we shall have 


F' :F :or /' = 


F'l 

F 



This follows from the fact that within the field of perfect elas¬ 
ticity the strains are proportional to the gradually applied forces 
or stresses. 

Substituting this value of /' in equation (26) we have 

2 F'l 

F' 2 ‘ + F 2 Fh + 2 F'l 

p ~ p’i ~ p’i .( 2 °) 

T 


F’H= 2 FVi + 2 F'Fl, 


F ' 2 -2 F'F = 


2FV1 


F' 2 —2F'F+F 2 = -~i — +F 2 , .... (29) 

F' 17 k 

P — ii \ ^ +1,.(30) 

F' . 

in which — is the ratio of the stress produced by the weight W 
falling through the height h, to the stress produced by the weight 















CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 23 


W acting as a gradually applied load. From equation (27) it 
is seen that this is also the ratio of the strains produced by the 
impact force W and the gradually applied force IF. 


PROBLEMS. 

9. A iooo-pound weight falls 2 inches before commencing to 

stretch the steel rod by which it is supported. This rod is 3 inches 
in diameter and 10 feet long. Find its elongation due to the 
falling weight. Ans. 0.048 inch. 

10. A steel rod 10 feet long and 1 inch square is to support 

a weight of 400 pounds which falls through a height h before it 
acts on the rod. Find the maximum value of h which will produce 
no permanent set in the rod. Ans. 5.98 inches. 


Application of the Principles of Longitudinal Stress. 

Elongation of a Rod of Uniform Cross-section.—In Fig. 3 
let the origin of coordinates be at B , the axis of Y vertical, and 
let D be any section at a distance y from B. Let it be required 
to determine the elongation of the rod due to its own weight, 
within the field of perfect elasticity. amnUMmnmmmmm nnurnmm 

Let DC = dy = length of an elementary por¬ 
tion of the rod, as CD ; ^ 

w = weight of a cubic inch of 
the rod in pounds; 

A = area of cross-section in 
square inches; 

L = original length of the rod in 
inches; 

/ = elongation of the rod in 
inches; 

F = tensile force in pounds on 
the area of cross-section 
at D. Then 

Ay — volume of rod between B and D, 

Awy = weight of rod between B and Z), 

F = Awy = force applied to cross-section D. 


W 


Y 

V/ 


B 


Fig. 3. 







24 


CIVIL ENGINEERING . 


Substituting in equation (5), transformed, 



FL 
EA ’ 


(3i) 


for i 7 its value above given, and for L , we have for the 

elongation of the elementary portion of the rod at D 



• • • (3 2 ) 


To find the elongation of the entire rod due to its own weight 
we must integrate this expression between the limits y — o and 
y=L. This gives for the elongation of the entire rod 

\wL 2 A 7 vL 2 

l ~ AE ~ 2E .^ 


If a weight wAL, the weight of the entire rod, is applied to 
the bottom of the rod, and the weight of the rod itself is neg¬ 
lected, we shall have for every cross-section F=wAL. Sub- 

stituting, in the equation l~, for F its value wAL, we have 

for the elongation of the entire rod within the field of perfect 
elasticity 

_ wL 2 

l =~F~ .(34) 


Comparing equations (33) and 34), we see that the elonga¬ 
tion oj a rod oj uniform cross-section due to its own weight is just 
one-half the elongation of the same rod caused by suspending 
from its lower end a weight equal to the weight of the rod , and 
neglecting the weight of the rod itself. 

The above discussion is equally applicable to the shortening 
of a vertical column under its own and a superimposed weight, 
within the field of perfect elasticity. 











CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 25 


PROBLEM. 

11. Determine the elongation of a vertical steel rod 1 inch 
in diameter and 50 feet long, under its own weight and a weight of 
20,000 pounds suspended from its lower extremity. 

Ans. 0.51 inch. 

Determination of Form of a Rod of Uniform Strength.—By a 

rod of uniform strength is meant one which has the same unit 
stress at every cross-section. 

Let ^ = unit tensile stress at any cross-section of rod, in pounds; 
F = applied tensile force in pounds; 

A = area of cross-section of rod in square inches. 

Then in the general expression 

F 

S e~ .(35) 

s e must be constant, and A must vary directly with F. 

Illustration.— A vertical rod is firmly jastened at its upper 
end and supports a weight at its lower end; let it he required to 
find the area oj cross-section at any point oj the rod if the weight 
0) the rod is considered. 

Let the axis of the rod shown in Fig. 4 be the axis of Y and 
the origin of coordinates at bottom. 

Let W = weight suspended from bottom of 
rod in pounds; 

w = weight of cubic inch of rod in 
pounds; 

L = length of rod in inches; 
y = distance in inches of any cross- 
section, as C, from bottom; 

A = area of the cross-section C in 

square inches; 

A'= area of the bottom cross-section B 
in square inches; 

^ 4 " = area of top cross-section A in 

square inches; 

W + w j A 3 y = total tensile stress on the cross-section C in pounds; 

5/'= allowable unit stress in pounds on any cross-section. 



I 

1 

Y 

w 

Fig. 4. 









26 


CIVIL ENGINEERING. 


Then 


W 4- w f A dy = s/'A .(36) 


In this equation A and y are the only variables; if we find 
an expression for A in terms of y, by assuming values of y between 
zero and L, we may determine the corresponding values of A 
and thus determine the dimensions of the rod. 

Differentiating the equation 


W+wj Aoy = s/'A . (37) 

with respect to A, we have 

wAdy = s e "dA .(38) 

Transposing we have 

wdy dA 

^7 r= T .( 39 ) 


Integrating we have 

wy 

= Nap. log A + C .(40) 

Making y = o, in which case A becomes equal to A', we have 

C= —Nap. log A' .(41) 


Substituting the value of C in (40) we have 

wy A 

—77 = Nap. log /l—Nap. log/l' = Nap. log-p. . (42) 

Passing to equivalent numbers we have • 

wy 

A = A'e s ‘", .( 43 ) 

in which e is the base of the Naperian system, or 2.718-I-. 








CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 27 

« 

Since the only force acting in the cross-section at the bottom 
is W, we must have 



W 




(44) 


Substituting this value of A' in (43) we have 


A = 


W e Se 


wy 

T» 




in which A is given in terms of y and known quantities. 

Making y = L, in which case A becomes equal to A”, we 
have 


A 


n 


wL 

W e^ 77 

c ff ‘ 


(46) 


Since the tensile force acting at the base is W=s,"A' and 
the tensile force acting at the top is 


W 4- weight of rod = s e "A", .... (47) 


the weight of the rod must be s e "(A" — A'). 

Form of Rod of Circular Cross-section.—No particular form 
has been assigned to the cross-section of the rod. Assume it 
to be circular and let 

r — radius of a section at a distance y from the bottom; 
r' = radius of bottom section; 
r" = radius of top section. 

Then 

A = 7 rr 2 , 

A' =7 :/ 2 


Substituting these values in equation (43) we have 


(48) 

(49) 










28 


CIVIL ENGINEERING. 


Cancelling n and taking the logarithm of both members, 


wy 

Nap. log r = Nap. log r' + —77.(51) 

2 S 


Making r = r" and y = L we have 

wL 

Nap. log r" = Nap. log r' + —77.(52) 

2S e 


In these equations we may substitute the common logarithm 


of r , r', and r" by applying the formula 


log a 


= Nap. log a. 


0.43429 

Equation (51) is the equation of a line whose rectangular 
coordinates are the values of r and y. It is the line cut out of 
the surface of the rod by a plane through its axis. 

The discussion above given applies equally well to a column 
of uniform strength to resist its own weight and a superimposed 
load. 


PROBLEM. 

12. A wrought-iron circular rod of uniform strength, 300 feet 
long, suspended vertically, supports its own weight and a weight 
of 50,000 pounds attached to its lower extremity. Determine 
the dimensions of the upper and lower cross-sections of the rod, 
and its weight. 

Ans. A'= 5 square inches; A" =5.524 square inches. 

Weight =5240 pounds. 

Determination of Thickness of Thin Cylinders and Spheres 
to Resist Internal Pressure.—The internal pressure of steam and 
water in pipes and spheres is usually considered as uniformly 
distributed over the surface, and the stress developed by the 
pressure to be uniformly distributed over the area of the section 
of rupture. 

Pipes.—If we consider a cylindrical water-pipe to be divided 
into two equal parts by a plane through its axis, it is evident 
that the resultant water pressure on one half will be equal to and 





CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 29 


directly opposed to the resultant pressure in the other half. The 
tendency of these equal internal pressures is to separate the pipe 
into two equal parts, by tearing it along two rectilinear elements 
at opposite extremities of the same diameter. 

Since, in each half-cylinder, the components of the normal 
pressure which are parallel to the plane of separation neutral¬ 
ize each other, the resultant pressure normal to the plane of 
separation is equal to the unit pressure on the surface, multi¬ 
plied by the area of the plane included between the rectilinear 
elements of rupture, or the plane whose width is the interior 
diameter of the pipe and whose length is the length of the pipe. 

Hence if p = internal pressure on a square inch of the pipe in 
pounds, 

/ = length of the pipe in inches, 
d = interior diameter of the pipe in inches, 
the two opposing resultant pressures normal to the plane of 
separation are each pdl. 

The tendency of the pipe to rupture is resisted by the tensile 
strength of the material in the plane of separation. 

If t = thickness of pipe in inches, 

2// = area of metal in the plane of separation in square 
inches, 

2S e tl = tensile stress on this area in pounds. 

For safety the total allowable stress in the metal must be 
equal to or greater than the normal pressure, or 

2 $ e "tl = or > pdl, . (53) 


in which s e " = allowable or safe unit tensile stress of the metal. 

To find the minimum allowable thickness of metal, we have, 
therefore, 


/ = 



(54) 


The allowable stress in water-pipes must be small because of 
the shocks, or water-hammer , due to the sudden stoppage of the 
flow of water in pipes when valves have been opened and are 
suddenly closed. 





30 


CIVIL ENGINEERING. 


Spheres.—A hollow sphere subjected to internal pressure 
tends to rupture along the circumference of a great circle. 

If p = internal pressure in pounds on the square inch of surface,. 
r = radius of sphere in inches, 

the resultant pressure in each hemisphere normal to the plane 
of separation is pzr 2 . 

The resistance of the metal in the plane of separation is the 
resistance of the ring of metal included between two concentric 
circles whose radii are r and r + t. The stress in the metal is 
therefore s e 7r [( r + 0 2 — ? ' 2 ] = s e 7i{2rt +1 2 ), 

in which is the unit stress. 

The least allowable thickness of metal is the value derived 


from the solution of the equation 

s e n n(2rt + t 2 ) —p 7 zr 2 .( 55 ) 

If t is very small in comparison with r, the second term in the 
parenthesis may be omitted. Hence 

2 S/'t = pr .( 56 ) 


The error thus committed is on the side of safety. 

In the discussions above, it is assumed that the stress due 
to interior pressure is uniform over the section of metal made by 
the plane of separation. This is not true when the metal is 
very thick, but is sensibly true for ordinary steam- and water-pipes. 

PROBLEMS. 

13. What should be the thickness of metal of an 8-inch cast- 
iron water-main to resist a water pressure of 300 pounds? 

Ans. 0.5 inch. 

14. A force of 500 pounds is applied to the piston-head of a 
forcing-pump which communicates its pressure to a hollow sphere 
10 inches in diameter. If the diameter of the piston-head is 1 inch, 
what should be the thickness of the metal of the sphere? 

Ans. 0.4 inch. 

Longitudinal Stresses Due to Changes of Temperature.— 

All materials employed in engineering practice expand and 




CLASSIFICATION OF STRESSES AND LONGITUDINAL STRESS. 3 1 


contract with changes of temperature. Within the field of per¬ 
fect elasticity, the elongations and contractions are directly 
proportional to the number of degrees of change. For each 
material, therefore, there is a constant, /', which represents 
its change per unit of length for each degree Fahrenheit from 
absolute zero temperature. If the ends of a rod are firmly fixed 
and it is then exposed to a change of temperature, it will be sub¬ 
jected to a longitudinal stress corresponding to the change of length 
it would have undergone had the ends been free to move. Repre¬ 
senting the length of the bar by L, its coefficient of linear expan¬ 
sion per degree Fahrenheit by /', the number of degrees change 
by n, and the total elongation due to n degrees by l, then 


l = j'nL, and the unit elongation is -j = j r n. . . (57) 


Since the unit stress is equal to the modulus of elasticity 
multiplied by the unit elongation, there results 

El 

s e or s c = j~=Ej'n, .(58) 


and the total stress in any cross-section 

s e A or s c A =AEj'n ,.(59) 

in which A is expressed in square inches. The values of /' for 
ordinary building materials are: 

Stone. 0.0000055 

Cast iron. 0.0000062 

Wrought iron. 0.0000067 

Steel. 0.0000067 


PROBLEM. 

15. A wrought-iron bar, 2 square inches in cross-section, has 
its ends fixed immovably between two blocks when the tempera¬ 
ture is at 6o° F. What pressure will be exerted on the blocks 
when the temperature is at 100° F.? Ans. 14,472 pounds. 







CHAPTER II. 


SHEARING AND TORSIONAL STRESS. 

9 

Simple Shearing Force.—If two thin flat bars of steel are 
laid one upon the other and thus fastened by rivets, any force 
applied to the bars to slide one along the other will be a simple 
shearing force as applied to the rivets. 

It is assumed from experiments— 

1. That all the fibers of each rivet suffer the same lateral 
distortion. 

2. That the cross-sections of the rivets which lie in the planes 
of the adjacent faces of the two plates move laterally in their 
own planes. 

3. That these effects are observed whatever be the form 
of the cross-section. 

It follows, therefore, since the material is homogeneous and 
all fibers offer the same resistance, that .the stress on the cross- 
section is uniformly distributed. 

If F = total force in pounds applied to distort the cross-section, 
A = area of cross-section in square inches, 
s s = shearing stress per square inch, 

then 

_F 

^ > • • • • • • * • (60) 

in which is the unit stress oj shear and equation (60) expresses 
the law of distribution of the stress over the plane of cross-section. 

The limit oj shearing elasticity oj a material is the greatest 
unit shearing jorce which may he applied to , or the greatest unit 
stress which may he developed in, the material without producing 
a permanent set. Unlike the limit of longitudinal elasticity, it 
cannot be directly measured, because of the impossibility of 
accurately measuring the distortions. 


32 


SHEARING AND TORSIONAL STRESS. 


33 


The coefficient oj shearing elasticity oj any material is the 
jorce or stress obtained by dividing any unit jorce or unit stress y 
not exceeding the limit oj shearing elasticity oj the material con¬ 
sidered , by the unit distortion produced by it. As the distortion 
cannot be accurately measured, the coefficient, like the limit, 
cannot be accurately determined. It is usually assumed to be 
equal to the coefficient of torsional elasticity. 

The modulus oj shearing , or the ultimate strength in shearing , 
is the unit shearing jorce or unit shearing stress at the moment oj 
rupture. This may be obtained from equation (60) by increas¬ 
ing F until rupture takes place. The value of at the moment 
of rupture is the modulus of shearing. It will be represented 
by V- 

The allowable or saje unit stress in simple shear is obtained 
by dividing the modulus oj shearing by a suitable jactor oj sajety. 
It will be represented by s s ". 


Constants in Pounds. 

Wood 

with 

Grain. 

Wood 

across 

Grain. 

Cast 

Iron. 

Wrought 

Iron. 

Steel. 

Modulus of shearing, s 8 . . . . 
Safe unit stress, s 8 " . 

600 

150 

3,000 

1,000 

20,000 

3. OOO 

40,000 

9,000 

50,000 

12,000 


Designing.—The ordinary problem of design is, as under 
tension and compression, to determine the area of cross-section 
of a piece to resist a given shearing force. 

Let F = the applied shearing force in pounds; 

A = required area of cross-section in square inches; 
s s " = safe unit stress in simple shear. Then 

7 .( 6l ) 

From this equation any one of the quantities can be deter¬ 
mined when all the others are known. 

PROBLEMS. 

16. What must be the least diameter of a steel bolt which is 
to resist a simple shearing force of 30,000 pounds? 

Ans. 1.78 inches. 


















34 


CIVIL ENGINEERING. 


17. What pressure is required to punch a hole f inch in diameter 

in a wrought-iron plate f inch thick? Ans. 29,452 pounds. 

18. What is the safe shearing load of a wrought-iron rivet 

| inch in diameter? Ans. 1768 pounds. 


Shear in a Rod or Column.—If a rod, as in Fig. 5, is sus¬ 
pended from a ceiling and supports at its bottom a force F, this 
force will produce tension in every horizontal cross-section of 
the rod. If, however, we assume any oblique section, as that 
shown in the figure, the force acting on this section may be 
divided into two components; one is perpendicular 
to the section and tends to tear the rod apart along 
the section, and the other is parallel to the section 
and tends to shear it off at the section. 

Let A= area of horizontal cross-section, AB; 

A'= area of oblique cross-section, BC; 

</> = angle A CB ; 

F = force applied to end of rod; 

T = component of F perpendicular to BC; 

S = component of F parallel to BC; 

T 

Fig - 5 - —=unit tensile force perpendicular to BC; 

S . , . , 

^p = unit shearing force parallel to BC. 

The weight of the rod itself is not considered. 

It is evident that the rod will have the greatest tendency 

T 

to tear apart along the plane on which is greatest, and the 



5 

greatest tendency to shear off along the plane on which -77 is 

A 

greatest. 

A T F 

From the figure T = F sin 6 , and A' = -—-; hence —=— s in 2 6 

sin 9 A A 

This is a maximum when sin 0 is a maximum, or 0 is 90°. 
The unit tensile stress is, therefore, greater on a horizontal 
plane of cross-section than on any oblique plane. From the 

figure S = F cos <j> and A'--—— hence -^- = ^- S * n ^ cos ^ 

S will therefore be a maximum when cos <£ = sin (f >, or <£ = 45°. 












SHEARING AND TORSIONAL STRESS. 


35 


-7 



A 

C 

G 


B 


5H 

->• 


5F 


The unit shearing stress on oblique planes is therefore greatest 
on the oblique plane, making an angle of 45 0 with the axis. 

Since the sine of 90° is unity, and the sine and cosine of 45 0 
are both Vo.5, the maximum unit stress of tension on the hori- 

F 

zontal cross-section is —, and the maximum shearing stress on 

F 

the oblique cross-section is 0.5 —. The rod will therefore rupture 

by tension along a horizontal cross-section unless the modulus 
of shearing is less than half the modulus of tension. If the grain 
of a wooden rod is not parallel to its axis, it may rupture along 
an oblique section. 

Shearing Stress in Plane Perpendicular to Plane of Cross- 
section.—Hitherto we have considered only the shearing stress 
in the plane of cross-section. If we study the sf 4 
deformation produced by a shearing force, it 5 4 
will be seen that there is also a shearing stress 
developed in planes normal to the first and nor¬ 
mal to the action line of the shearing force. 

In Fig. 6 let ABCDEFG be an elementary 
molecule in the interior of the rivet subjected to simple shear. 
The faces BA E and DCFG are parallel to the plates united 
by the rivet and hence parallel to the applied shearing force. 
Let 

dd = AB , the depth of the molecule; 
db = AE, “ width “ “ 

dl=AC, “ length “ “ “ 

dH and dH = elementary shearing force developed on faces 

AEFC and BDG; 

dF and oF = elementary shearing force applied to the face 

CDGF and the resistance developed by it in 
the face ABE attached to its adjacent mole¬ 
cule ; 

(s s )i =unit stress on each face odob; 

(s s )iddob = total stress on each face odob ? 

Hence 

dF = ( s s )iddob . 

The effect of the force dF is to move the face on which it 
acts, in its own plane, and to increase the angles of the elemen- 


Fig. 6. 









36 


CIVIL ENGINEERING. 


tary molecule at B and C, and decrease those at A and D. This 
will cause the planes AEFC and BDG to slide on the correspond¬ 
ing faces of the molecules above and below and thus produce a 
shear in the upper and lower faces of the elementary molecule. 
Since the molecule is at rest and the vertical shearing forces 
form a couple, the horizontal shearing forces must also form a 
couple. 

Let (s s ) 2 = unit shear on the upper face. 

Then (s s ) 2 dbol = total shear on the upper face, 

and oH = (s s ) 2 obdl. 

The same expression represents the equal shear on the lower 
face which has the same area. 

There will be no shear on the faces A BCD and FFG, since 
the molecules on either side are each subjected to a couple equal 
to dF, dF. 

The molecule ABCDEFG being in a state of rest under 
the action of the two couples dF and dF, oH and oH, the moments 
of these couples must be numerically equal. Hence 

oFXdl = dHxdd . (61) 

Substituting for dF and oH their values we have 

(s s )idbddol= (s s ) 2 dbdldd .(62) 

Hence ($ a ) 1 = (s s ) 2 , or the unit shearing stresses on the vertical 
and horizontal jaces oj the elementary molecule are equal. 

Torsion. 

Torsional Moment.—Any force whose action line lies in a 
plane normal to the axis of a shaft and does not intersect that 
axis, will produce torsional stress in the shaft, unless the shaft 
is absolutely free to rotate on its axis. The moment of 
the torsional force with respect to the axis is called the torsional 
moment, and is positive if it tends to rotate its plane clockwise, 
and negative if it tends to rotate its plane counter-clockwise. 

If a shaft, as in Fig. 7, is fixed at one end, and at the other 
is subjected to a torsional bending force F whose moment with 


SHEARING AND TORSIONAL STRESS. 


37 


respect to the axis ab is positive, it is evident that this tor¬ 
sional moment in the end cross-section will be transmitted from 
cross-section to cross-section 
until it reaches the fixed end. 

Hence every cross-section be¬ 
tween the applied force and the 
fixed end will be subjected to 
the same torsional moment. If 
a second torsional force F is ap¬ 
plied at the middle point of the 
shaft so as to produce a posi¬ 
tive torsional moment equal to 
the first, it is evident that the resultant torsional moment between 
the fixed section and the middle point will be doubled; the 
moment between the forces will, however, remain unchanged. If 
the second force is applied so as to produce a negative torsional 
moment, the resultant torsional moment between the second force 
and the fixed section will be reduced to zero, while that between 
the forces again remains unchanged. Hence 

The torsional moment at any cross-section of a shaft acted 
upon by torsional forces is equal to the algebraic sum of the tor¬ 
sional moments of all the forces acting on one side of the section. 

In the above discussion the bending effect of the force F 
has not been considered, since it is independent of the torsional 
effect. The bending effect will be considered in the chapters 
following. 

Torsional Stress.—The theory of torsion is based on the 
following hypotheses, derived from observation: 

1. The effect of a torsional force is to rotate each cross- 
section, between the point of application and the fixed section, 
about the axis of the shaft. 

2. Each cross-section remains a plane surface during its rotation. 

3. The amount of rotation of any section will vary with its 
distance from the fixed section. 

The effect of the torsional moment of F, Fig. 7, is there¬ 
fore to turn the end section and all the other cross-sections with 
it around the axis until the resultant moment of resistance 
developed in the fixed section is equal to the torsional moment; 
this prevents any further motion in its adjacent section. The 









38 


CIVIL ENGINEERING. 


resultant moment of resistance developed in this adjacent section 
will, when it becomes equal to the torsional moment, prevent 
further rotation of the section to its right. This effect will be 
transmitted from section to section until it reaches the plane of the 
force F. The fiber ej, which was originally straight, will become 
a spiral ed. The angle jad is called the angle oj torsion , and the 
arc dj the arc oj torsion. The amount of distortion of the mole¬ 
cules of the fiber ej will depend upon their distances from the 
fixed section. 

Law of Distribution.—If we examine the end section, it is 
seen that the fiber at a is not distorted, and the fiber at / is dis¬ 
torted through the arc dj. The distortion of any other fiber along 
the radius aj is distorted an amount dependent on its distance 
from a. Since within the elastic limit the stress is proportional 
to the distortion, the stress on any jiber oj the cross-section will 
vary with its distance jrom a, the axis oj rotation. 

Unit Stress.—Let R = radius of shaft in inches; 

r = variable distance in inches measured 
along the radius; 

dyoz = elementary area of cross-section; 
s/” = unit torsional stress in pounds in a fiber 
at a unit distance from the axis; 
s* = unit torsional stress in pounds in the 
surface fiber. Then 

s”'r= unit torsional stress in a fiber at a distance r 
from axis; 

s t '”dydzr = total stress in an elementary area at a distance 
r from the axis; 

s/"dydzr 2 = moment of s"'dydzr about the axis; 

s t” f J dydzr 2 = resultant moment of all the stresses about the axis, 

or the torsional moment oj resistance oj the 
cross-section. 

As stated above, the rotation of any section will cease when the 
torsional moment of resistance developed in the section becomes 
equal to the torsional moment of the twisting force at the section. 
Whence 

St" f f SySzr 2 = M t = Fr, . (63) 

in which M*=Torsional moment. 


SHEARING AND TORSIONAL STRESS. 


39 


The expression / I dydzr 2 occurs in mechanics as the moment 

of inertia of a plane area about an axis through its center and 
normal to its plane, or the polar moment oj inertia oj the cross- 
section. In this case it is the moment of inertia of the cross- 
section of the shaft about the axis of the shaft. 

Representing this moment by I p and substituting in equation 
(63) we have 

s t '"I P =M h .(64) 

in which s/"I p = moment of resistance, from which 


M t 

$/"=— = unit stress at a unit distance from the axis. 

Multiplying both members by r we have 
M t r '• . 


s/"r= 


unit stress at a distance r from the axis. (65) 


Since the stress is not uniformly distributed over the area 
of cross-section, there can be no unit area, 1 square inch, over 
which the stress is uniform, as in tension, compression, and 
shearing. The unit stress at a distance r from the axis is, there¬ 
fore, the stress on a hypothetical area 1 square inch in cross- 
section, each element of which is subjected to the same stress 
as the elementary area dydz at a distance r from the axis. 

Limit of Torsional Elasticity.—If we substitute in equation 
(65) R for r } we shall have for the surface fiber 

s/"R=s t = ~ .( 66 ) 


If the force F in Fig. 7 is gradually increased from zero, 
we shall eventually reach a value which will produce a perma¬ 
nent set in the surface fiber. This limiting value of s t is the 
limit oj elasticity oj the material in torsion or the limit oj torsional 
elasticity oj the material. 

The limit oj torsional elasticity oj a material is the greatest 
unit torsional stress which can he developed in the surface fiber oj 
a shaft oj the material without producing a permanent set. 







40 


CIVIL ENGINEERING. 


Coefficient of Torsional Elasticity. —Within the field of per¬ 
fect elasticity the unit stress must be proportional to the unit 
distortion. 

If En == coefficient of torsional or lateral elasticity, 

/ = distortion of surface fiber = fd in Fig. 7, 

/' = distortion of fiber at unit distance from axis. 

L = length of every fiber = ef in Fig. 7, then 

E t l E t VR 
St ~ L~ L ' 

and 

Et - 7-.(67) 

L 

In this expression l is expressed in linear units; it may be 

l 180 

expressed in degrees by multiplying — by . The coefficient 

oj lateral elasticity 0f any material is the quantity obtained by 
dividing the unit torsional stress on the surface fiber of a shaft , by 
the unit distortion of that fiber, within the field of perfect elasticity. 

Modulus of Torsion. —If, in equation (66), M t be increased 
until the shaft breaks, the corresponding value of s t will be the 
modulus of torsion of the material of the shaft or its ultimate 
strength in torsion. It will be represented by sf. The modulus 
of torsion is the unit torsional stress on the extreme fiber of a 
shaft of the material , at the instant of rupture. 

Safe Unit Stress. —The greatest unit stress to which it is 
deemed advisable to subject the surface fiber is called the safe 
or allowable unit stress in torsion. It will be represented by s t ". 

The values of the constants of torsion for ordinary building 
materials are given approximately in the following table: 


Constants. 

Wood. 

Cast Iron. 

Steel. 

Wrought 

Iron. 

Coefficient of elasticity, Et . . 

Modulus of torsion, s/ . 

Safe unit stress, st" . 

1,500 

375 

6,000,000 

25,000 

4,000 

12,000,000 

60,000 

10,000 

10,000,000 

50,000 

9,000 



Designing.—The problem of designing a piece to resist tor¬ 
sional stress is usually that of determining the diameter of a 





















SHEARING AND TORSIONAL STRESS. 


41 


shaft to resist a given torsional moment, or conversely, to deter¬ 
mine the torsional moment which can be resisted by a given 
shaft. 


The shaft is usually of circular cross-section, though the 
law of distribution is approximately true for any cross-section 
whose form is that of a regular polygon, as a square, a regular 
hexagon, etc. 

If, in equation (66), we substitute for s t its safe value s t ", 
we shall have 


s" = 


M t R 

I P ’ 


( 68 ) 


or 


R = 


M t 


(69) 


in which M t = torsional moment at the cross-section in (inch- 

pounds) ; 

= polar moment of inertia of the cross-section ex¬ 
pressed in terms of R; 

R = radius in inches of the circular cross-section, or 
the radius in inches of the circumscribed circle 
of the regular cross-section; 
s t " = safe unit stress in pounds of the material in tor¬ 
sion. 

From these equations we can determine the value of any 
one of the quantities when all the others are given. 


PROBLEMS. 

19. A circular steel shaft is subjected to a torsional moment 
of 10,000 (foot-pounds). What should be its diameter? I P =^7 zR A . 

Ans. Diameter 3.94 inches. 

20. What torsional moment in (foot-pounds) may be safely 
borne by a steel shaft six inches in diameter? 

Ans. 35,343 (foot-pounds). 

21. A circular steel shaft is 6 inches in diameter and 20 feet 
long; through what angle may it be safely twisted? 

Ans. 3.8 degrees. 

Dangerous Section.—From the equation 



M t R 

h 


(7°) 








42 


CIVIL ENGINEERING. 


it is seen that for any shaft of uniform cross-section s t must 
vary directly with M t , since R and I p are constant. Hence the 
greatest fiber stress in the surface fiber will take place at the 
section of greatest torsional moment. As this is the section 
where the shaft of uniform diameter is most liable to break, it 
may be called the dangerous section. 

Shaft of Uniform Strength.—In equation 



M t R 

Ie 



the safe unit stress s t " is a constant. If the condition is imposed 
that the maximum stress or the stress in the surface fiber in 
each cross-section shall be s t ", the shaft becomes one of uniform 


strength. In such a shaft the value of the ratio of M t to j- must 

p 

be constant. 

If the cross-section is circular, I p = %t:R 4 ; hence the ratio 
jr = \-R*. In a circular shaft of uniform strength, therefore, 


R must vary directly with 



Elastic Torsional Resilience.—The elastic torsional resilience 
is the elastic energy stored in a shaft by twisting it within its 
field of perfect elasticity. It is equal to the work done by the 
twisting force. Conceive a scale-pan suspended from a tape 
wrapped around the free end of a horizontal shaft which is fixed 
at the other end and supported throughout to prevent bending. 
By gradually introducing weights into the pan the shaft will 
be brought under stress and twisted, until for a particular weight, 
W, the stress in the surface fiber is s t , and the angle of torsion 
n degrees. During the process of loading, the scale-pan will 
descend a distance equal to the arc of torsion at the circum- 
m:R . nr.R 

ference, or Representing by l and remembering 

that the work of a gradually applied force is equal to one-half 
its intensity into its path, we have for the work of the force 


work = 


Wl 


2 


• (72) 











shearing and torsional stress. 


43 


In this expression if we substitute for l its value Rl', in which 
l' is the length of the arc of torsion at a unit distance from the 
axis, we shall have 


work = resilience = 


IV Rl' M t l' 
2 2 


( 73 ) 


To obtain an expression for the resilience in terms of the 
stress we must substitute for M t and l' their value deduced from 
the expressions 

M t = ^~ and /' = .(74) 

Making these substitutions we have 


work = resilience = 


1 s t I. 


R 


s t L s t 2 IpL 
E t R~ 2 E t ' R 2 • ’ 



Substituting for I p its value Ar p 2 , in which A =area and = its 
radius of gyration, 


work = resilience = 


2 E t 


y 2 

LP-.J T 

R2 



This expression becomes the maximum elastic torsional resili- 

s 2 

ence when s t becomes the limit of torsional elasticity, and 

2 Hj t 

becomes the modulus of elastic torsional resilience. 

The modulus 0j torsional resilience oj any material is the quan¬ 
tity obtained by dividing the square oj its limit oj torsional elasticity 
by twice its coefficient oj lateral elasticity. It measures the capacity 
of that material in comparison with other materials to store up 
the work expended on similar shafts by the same torsional force, 
within the field of perfect elasticity. 

Power Transmitted by Shafts.—Conceive a circular shaft 
with a wheel at each end for a driving-belt; the belt at the end A 
being connected with the engine, and that at B with a machine. 
It is assumed that neither belt slips. 

Let the uniform pull on belt A in pounds be represented by 
F, and the radius of the wheel in inches by r. At each revolution 
of the wheel the path of the force F is 2nr, and in n revolutions 
it will be 2T,rn. The work performed by the force F in making 













44 


CIVIL ENGINEERING. 


n revolutions will be 2izrnF. If we assume that n revolutions 
are performed in one minute, this work per minute may be reduced 
to horse-power by dividing by 33,000X12, a horse-power being 
33,000 foot-pounds per minute. Hence 


Horse-power in inch-pounds = 


27 :rnF 
33,oooX 12’ 



In this equation there are four variable quantities, the horse¬ 
power, n, F, and r. This enables us to make assumptions on any 
three and deduce the corresponding value of the fourth. 

Since 

Fr=M t = S ~f'i 


we may, by substituting, obtain an expression for the horse-power 

in terms of s t and ~, or 

K 


Horse-power 


27 znsj p 

* • 

33,000 X 12 X.K’ 



or substituting for F p its value in terms of R, ^R 4 , 


Horse-power 


s t - 2 R 3 n g. 8 ys f R 3 n 
396,000 396,000 


s t R 3 n 

-^ qo , approximately. (79) 


In this equation there are also four variable quantities, the 
horse-power, the unit stress in the extreme fiber, the radius of 
the shaft, and the number of revolutions. This enables us to 
make assumptions on any three and deduce the corresponding 
value of the fourth. 


PROBLEMS. 

22. Find the diameter of a steel shaft which will safely trans¬ 
mit 100 H.P. when making 225 revolutions per minute. 

Ans. 2.43 inches. 

23. Find the horse-power which can be safely transmitted by 

a hollow steel shaft (R = 10 inches, r — 5 inches) when making 100 
revolutions per minute. I p =%x(R*-r 4 ). Ans. 7438 H.P. 








CHAPTER III. 


FLEXURE OR BENDING. 


Bending Moment.—A simple bending force is a force whose 
action line is perpendicular to and intersects the axis of a beam. 
Its effect is to bend a beam upon which it acts. 

The bending moment of such a force with respect to any cross- 
section of a beam is the product of the intensity of the force, by 
its lever-arm or the distance between the action line of the force 
and the plane of cross-section. It is usually expressed in inch- 
pounds. To distinguish it from inch-pounds of work it will be 
represented (inch-pounds). 

In all our discussions we have assumed clockwise moments 
positive and counter-clockwise moments negative. Bending 
moments are, however, considered positive when they produce 
compression in the upper fibers of the beam and tension in the 
lower fibers. The two systems of notation will agree if the center 
of moments is to the right of the forces; if the center of moments 
is to the left of the forces, we must F >> 

give the positive sign to counter- a 
clockwise moments. ~ 

Let Fig. 8 represent a beam in a F . 
state of rest acted upon by four bend¬ 
ing forces F', F", F'", F l \ Let c be 
a cross-section of the beam. Then will 


1 d 


Fig. 8. 


F' Xac = positive moment and positive bending moment; 
F" Xbc = negative “ “ negative 

F'" Xdc = positive “ 

F IV Xec = negative “ “ positive 


< < 


i i 


t < 


< < 
< i 
11 


The bending moment at c in Fig. 8 is 


M=+F'Xac-F"Xbc= -F'" Xdc+F iy Xec. . . (8o) 

45 


I 





46 


CIVIL ENGINEERING. 


Hence 

The bending moment at any section oj a beam is the bending 
moment oj the resultant oj all the bending forces acting on either 
side oj the section , or the algebraic sum oj the bending moments oj 
all the bending forces acting on either side oj the section. 

The determination of the bending moment at any section of 
a beam therefore consists in— 

1. Finding the intensities and action lines of all the bending 
forces which act on one side of the section. This may be done 
by employing the equations of equilibrium, as modified for ver¬ 
tical parallel forces. 

2. Finding the resultant moment of such forces with respect 
to the section considered. 

It is evident from the explanations above given that it will 
usually be better to consider the forces on the left of the section, 
since the moment of each force and its bending moment will 
then have the same sign; if, however, merely a numerical value 
of the moment is desired, or the forces on the right can be more 
readily determined, the bending moment may be deduced from 
the forces on the right, if proper attention is paid to the signs. 

Vertical or Transverse Shear.—A second effect of bending 
force is to move the consecutive planes of cross-section along 
each other and thus develop shearing stress. The total effect 
at any cross-section is called the vertical or transverse shear , or 
simply the shear at that cross-section. As beams are usually 
horizontal, the term vertical shear will be used, and it will be 
represented by V s . 

If we consider the cross-section c in the beam, Fig. 8, to be 
at rest, all the vertical forces must be transmitted to it through 
the internal resistances of the beam itself. 

Thus the force F', acting in the end section, will tend to shear 
off that section, but it will be opposed by the resistance to shear¬ 
ing of the fibers connecting it with the next section. The effect 
of F' will therefore be transmitted from section to section, from 
a to c. In a similar manner the effect of F", acting in a contrarv 
direction, will be transmitted from section to section, from b 
to c. 

The resultant shear, or force transmitted to c, will be their 
difference. 


FLEXURE OR BENDING. 


47 


Similarly on the right of c , the force transmitted to c y or the 
shear at c, is the difference between F'" and F 1V . Since the 
section c is at rest, the force transmitted to it from the left must 
be equal in intensity but contrary direction to that transmitted 
to it from the right. 

The shear at any section is therejore equal to the resultant oj 
all the bending forces acting on either side oj the section. 

Since the bending moment is the resultant of all the bending 
forces acting on one side of the section, multiplied by its distance 
from that section, we have 


and 


M—V s x, .... 

. . . (81) 

m 

. . . (82) 

v* 

ox ’ 


in which M — the bending moment at any section; 

F s ^the vertical shear at same section, oy resultant of 
vertical forces on one side of the section; 
x— lever-arm of the resultant of vertical forces on one 
side of the section with respect to that section. 

The vertical shear therefore measures the rate oj change oj the 
bending moment. 

The shear at every section in that part of the beam in which 
the bending moment is an increasing algebraic function of x is 
considered positive; the shear at every section in that part of 
the beam in which the bending moment is a decreasing function 
of x is considered negative. 

If the shear at any section is determined by finding the result¬ 
ant of the forces on the left of the sec- -hf . 

tion, the shear will be positive if this 
resultant acts upwards, and negative 
if this resultant acts downwards. If 
only the numerical value of the shear 
at the section is desired, its sign is 
immaterial. 

Theory of Bending or Flexure- 

Let Fig. 9 represent a cantilever beam fixed in a horizontal posi¬ 
tion and acted upon by a single bending force F at its outward 


7 

i 



+ h 

ZZZ— J 

< " H 

i 


1 

y r 


-F 


Fig. 9. 
















48 


r 


CIVIL ENGINEERING. 


extremity. Let it be required to determine the law of distribu¬ 
tion of the stress over any cross-section, as that at c. 

Laws of Distribution.—The common theory of flexure is 
based upon the following hypotheses derived from experiments 
on the effect of bending forces on a beam. The forces are 
assumed as acting in the vertical plane through the axis of the 
beam. 

1. The fibers on the convex side are lengthened , those on the 
concave side are shortened; the stresses on the separate fibers are 
therefore longitudinal and either tensile or compressive. 

2. Between the lengthened and shortened fibers there is a neutral 
surface whose fibers are neither lengthened nor shortened. The 
line cut out of this surface by the vertical plane through the axis 
is called the mean or neutral fiber , or the elastic curve. 

3. The planes of cross-section normal to the fibers before bend¬ 
ing will remain planes and be normal to the fibers after bending. 

4. The planes of cross-section have no motion of translation 
along the neutral fiber, but simply rotate about the line cut out of 
the neutral surface by the plane of cross-section. This line is 
called the neutral axis. 

In Fig. 9, MN is the mean fiber, and the horizontal line, 
perpendicular to it at N , is the neutral axis of the section cc. 

Under the action of the force F, therefore, the cross-section c 
will rotate about its neutral axis N, and the fibers above the 
axis will be lengthened and those below will be shortened. The 
strain or stress on any fiber will therefore vary directly with its 
distance from the neutral axis. 

Unit Stress.—In Fig. 9 let 

F = intensity of bending force in pounds; 

/ = lever-arm of F with respect to section c, in inches; 

M = —Fl = bending moment at section cc in (inch-pounds); 

+ H = resultant tensile stress in fibers above the neutral axis; 

—H = resultant compressive stress in fibers below the 
neutral axis; (these resultants must be equal to 
each other, since by hypothesis the section has no 
motion of translation along the axis of the beam.) 
vertical distance between the resultant tensile and 
compressive stresses or the lever-arm of the couple 
HH. 


FLEXURE OR BENDING. 


49 


If the beam is separated at the section cc, and longitudinal 
stresses in the fibers at the section replaced by the forces +H 
and —H, and the vertical shear by the force +F acting upwards, 
the segment between the section and the free end will be in a 
state of rest, and the four forces acting on it will be in equilib¬ 
rium, or 


H(h=Fl = M. 

To determine an expression for Hd i in terms of the unit 
stress, let 

s'" = unit longitudinal stress at a unit distance from 
the neutral axis; 

5"'y = unit longitudinal stress at a distance y from the 
neutral axis; 

dydz = elementary area of cross-section at c ; 
s'"ydydz = stress on the elementary area; 
s'"y 2 dydz = moment of stress on the elementary area with 
respect to the neutral axis; 

s'" j jy 2 dydz = moment of stress on the entire area of cross-section 

with respect to the neutral axis, if integrated 
between proper limits. Hence 

s'" f J y 2 dydz = Hdi .(83) 


The expression J J y 2 dydz occurs in mechanics as the moment 

o) inertia oj a plane area about an axis in its own plane; in this 
discussion it is the moment of inertia of the area of the cross- 
section cc about the neutral axis. If we represent this moment 
by I, we have 

s'" I = Hd\ =Fl = M .(84) 

In this equation the moment Hdi or its equivalent s'"I is 
called the moment of resistance of the fibers. 

From equation (84) we have 


5° 


CIVIL ENGINEERING. 


s'" = y-, for the unit longitudinal stress at a unit distance from 
the neutral axis; 

s ,f, y = ^Y~ } for unit longitudinal stress at a distance y from the 


neutral axis. 

Position of the Neutral Axis.—Let 

5'"^ = unit longitudinal stress at a distance y from the 
neutral axis, 

dydz = elementary area of the cross-section cc, 
s'"ydydz = stress on the elementary area at a distance y from 
the neutral axis, 


s'" J J ydydz = total longitudinal stress on the area of cross-section 
• ' 

cc, if integrated between proper limits. 

The expression J Jydydz is called the static moment oj an 


area about its axis oj rotation, and is equal to the total area mul¬ 
tiplied by the distance of its center of gravity from the axis of 
rotation, or 


/ fyoyoz=Ad 2 , .(85) 


in which A= area of cross-section in square inches; 

d 2 = distance of its center of gravity from the axis of 
' rotation in inches. 

In our discussion A is the area of the cross-section cc, and 
d 2 is the distance of the center of gravity of the cross-section from 
the neutral axis. 

Since +H, the resultant stress in the fibers above the neutral 
axis, is equal to —H, the resultant stress in the fibers below the 
neutral axis, but acts in a contrary direction, the resultant hori¬ 
zontal stress in the cross-section must be zero. 

Hence in the section cc 


s'" J f ydydz = s'"Ad 2 = o, 



Since neither s'" nor A is equal to zero, d 2 must be equal 
to zero, or the neutral axis must pass through the center oj gravity 
oj the cross-section. 










FLEXURE OR BENDING. 


5 1 


Maximum Unit Stress in Cross-section.—From the equation 


* 



we may determine the maximum unit stress in the cross-section 
whose resultant bending moment is M by substituting for s n 'y, s, 
the unit stress in the fiber farthest from the neutral axis, and 
for y, y r , the distance of the extreme fiber from the neutral axis; or 


MV M 



( 88 ) 


or s(^J=M, .(89) 

in which —7 is the section modulus . 

y 

The stress in the surjace fiber oj a beam at any cross-section 
is equal to bending moment at the section divided by the section 
modulus oj the section. 

Modulus of Flexure. — If M in the second member of 
equation (88) is increased without changing the section modulus, 
j will be increased, and finally the extreme fiber will rupture. 
The value of s at the moment of rupture, represented by s', is 
called the modulus oj flexure , or the ultimate strength in flexure. 

The modulus oj flexure is therejore the unit longitudinal stress 
on the extreme fiber oj a beam at the moment oj rupture by a 
bending jorce. 

Allowable or Safe Unit Stress. —The allowable or saje unit 
stress, s ", is the greatest unit longitudinal stress to which it is 
advisable to expose the fiber farthest from the neutral axis. 

For safety we must have 

MV , x 

s" = or > '-j-' .(90 \ 


at every cross-section of the beam. 










5 2 


CIVIL ENGINEERING. 


The general values of these constants for ordinary build¬ 
ing materials in pounds per square inch are: 


Constants. 

Wood. 

Cast Iron. 

Steel. 

Wrt. Iron. 

Modulus of flexure.. . .. 

6, 000 
1,000 

36,000 

6,000 

60,000 

15,000 

48,000 

12,000 

Safe unit stress. 



Factor of Safety.—The factor of safety of any material is 
the ratio of the ultimate to the safe unit stress under the assump¬ 
tion that this ratio is the same for tension, compression, shear¬ 
ing, torsion, and bending. It may be employed to determine 
the safe unit stress when the tables give only the ultimate or 
breaking unit stress. 

The factor of safety employed depends on the uniformity 
of structure of the material and the character of the force which 
must be resisted. 

The following table gives safe values of such factors for 
ordinary building materials: 



Quiescent 

Force, 

Buildings. 

Varying 

Force, 

Bridges. 

Shocks, 

Machines. 

Stone and brick. 

15 

23 

3 ° 

Timber and cast iron. 

8 

12 

16 

Steel and wrought iron. . .. 

4 

6 

8 


Section Modulus.—The section modulus ^7 is the quotient 

resulting from dividing the moment of inertia of the cross-sec¬ 
tion about the neutral axis, by the distance of the extreme sur¬ 
face fiber from that axis. 

The value of I for any form of area of cross-section is deter¬ 
mined by integrating the expression f jy 2 dydz between the 

proper limits of y and z. In this expression z is the coordinate 
of any elementary area measured parallel to the neutral axis, 
and y is the coordinate of the same area measured in a direc¬ 
tion perpendicular to that axis. 

The value of y' is the greatest distance of any fiber from 
the neutral axis. 


























FLEXURE OR BENDING. 


53 


The neutral axis, it will be remembered, always passes through 
the center of gravity of the cross-section. 

The values of /, /, and -7 for ordinary forms of cross-sec¬ 


tion are tabulated in engineering manuals. 1 Some of these are 
given in the following table: 


Forms of Cross-section. 

Dimensions. 

/ 

y 

Section 

Modulus, 

~y’ 

Rectangle. . . .. 

f breadth b 
[ depth d 

bd 3 

d 

bd 2 


12 

2 

6 

Square with a vertical 
diagonal.. 

f breadth d 
\ depth d 

d 4 

<tv7 

d 3 

8-5 


I Z 


Triangle with horizon¬ 
tal base. 

f base b 

\ altitude d 

bd 3 

36 

§d 

bd 2 



24 

P i rrl f* . 

diameter d 

ltd 4 

d 

d 3 


64 

2 

10 + 

Ellipse ... 

f axes b and d 

Tlbd» 

d 

bd 2 

I beam, cut from rect¬ 
angle b X d } by remov¬ 
ing two areas, each 
d'Xb' 

\b = neutral axis 

64 

(bd'-b'd'*) 

2 

d 

10 + 

(bd*-b'd't) 

2 


12 

2 

6 d 


It will be observed that as a general rule — is the function of 

the two dimensions b and d; in the circle and square it is a func¬ 
tion of a single dimension d or r. 

Design.—The problems m design are either to find the dimen¬ 
sions of a beam which will support a given load, or to find the 
load which will be supported by a beam of given dimensions. 

Beams may be divided into two general classes: beams oj 
uniform cross-section and beams oj uniform strength. 

The first is the ordinary form of beam, as it is a form easily 
sawed out of wood or rolled out of steel or wrought iron. 

The second, or an approximation to it, is employed when it 
is desired to reduce the weight of the material in a beam to a 


Cambria handbook, pp. 144--151. 





























54 


CIVIL ENGINEERING. 


minimum. Built-up beams of wood or metal plate are often 
made of this form; formerly when cast-iron beams were used 
they were also of this form. 

Beams of Uniform Cross-section.—In beams of uniform cross- 
section the section modulus is the same at every cross-section. 
Therefore, in the general equation 

f-M, .(91) 

s, or the unit stress in the surface fiber, varies directly with M and 
is numerically greatest in the cross-section where the bending 
moment is greatest. As the beam is more liable to break at this 
section than at any other, it is called the dangerous section. If 
M m = the resultant bending moment at the dangerous section in 
(inch-pounds), 

-7 = section modulus, 

/ 

s" = safe unit stress in pounds, 

then the equation of condition for safety in a beam of uniform 
cross-section is 

s"I 

or >M„ .(92) 

If the problem is to determine the dimensions of a beam 
to carry a given load, the value of s" is taken from the table; 
M m is computed from the loading, as will be explained hereafter; 

and is expressed in terms of b and d. As all the quantities 

in the equation, except b and d, are known, we may assume one 
of the dimensions and compute the other. 

Let it be required to find the minimum depth of a wooden 
beam of uniform cross-section, whose width is 6 inches, to 
resist a bending moment of 100,000 (inch-pounds) at the danger¬ 
ous section. 

From the table 

s” = 1000, 






FLEXURE OR BENDING . 


55 


hence ioood 2 = 100,000, 

d 2 = 100, 
d== 10 inches. 

We may therefore always determine the dimensions of the 
cross-section of a beam of uniform cross-section when we know 
the section modulus in terms of b and d, and the bending moment 
at the dangerous section. 

Beams of Uniform Strength.—In beams of uniform strength 
the unit stress in the surface fiber at every cross-section is the 
same; or in the general equation 

j=M .( 93 ) 

5 is the same for every cross-section and -7 must vary directly 
with M. 

For safety^ should be equal to or less than the safe unit 
stress, or 

,, M 

s" = or >—..( 94 ) 

7 

If we substitute for -7 its value in terms of b and d , as, for 

/ 

instance, in a rectangular cross-section, we have 


s" = 


6 M 

bd 2 


( 95 ) 


In this equation we may assume a value tor one of the dimen¬ 
sions, as d, and we shall have 


s"d 2 M 
~~ 6 ~ = ~b 


( 96 ) 


In this equation the first term is constant, hence at every 
section of the beam the value of b must be such that the ratio 

— is constant and equal to the first member of the equation. 
b 









56 


CIVIL ENGINEERING. 


Since the bending moment varies from point to point of the 
axis of a beam, it follows that the dimensions of the cross-section 
of a beam of uniform strength can only be determined when we 
know the law governing this variation. 

The Law of Variation of the Bending Moment and the 
Bending Moment at the Dangerous Section.—As the resultant 
bending moment at any section of a beam is the resultant moment 



of all the bending forces acting on one side of the section, it is 
dependent on the intensities and action lines of the applied forces, 
or on the method of loading and supporting the beam, and is 
independent of the form and dimensions of cross-section. 

According to the method of loading and supporting we have, 
in Fig. io, the following classes of beams: cantilevers, i and 2; 













































































































FLEXURE OR BENDING. 


57 


beams resting on two end supports, 3 and 4; beams fixed at one 
end and supported at the other end, 5 and 6; beams fastened at 
both ends, 7 and 8. 

To prevent confusion, in deducing the law of variation of the 
bending moment, only the axis of the beam will be represented 
in the figures. 

1 . Cantilever without Weight Supporting a Load at its 
Extremity. —In Fig. n let 


0 = 
0 X = 

OM = 

ov s = 

l = 

w = 

c = 


origin of coordinates; 
axis of beam and axis 
of x; 

axis of moments; 

= axis of shears; 
length of beam 
inches; 

load concentrated 
end in pounds; 
any section of the beam 
between O and X ; 
bending moment at c; 


M 1 V, 


in 


at 



Fig. 11. 


M 

= bending moment at dangerous section. 


m 


Considering the forces to the left of c, we have from definition 
the bending moment at c , 

M=-Wx .(97) 


This equation, called the equation 0 ) bending moments , ex¬ 
presses the law of variation of the, value of the bending moment 
from section to section. Since M is a numerically increasing 
function of x, it is greatest numerically at X, where x is greatest 
and equal to /; and least numerically at O, where x=o. The 
dangerous section is therefore at X, and 

M m = —Wl .(98) 

If we lay off XA=lVl, the line (1) will be the line whose 
equation is M=-Wx. This line is called the line or curve of 
bending moments. From it bv inspection we may determine 
the position of the dangerous section. 










53 


CIVIL ENGINEERING. 


Shear .—By definition, the vertical shear at any section of the 
beam, as c , is equal to — IF. It is negative because the force is 
to the left of the section and acts downwards. Hence 


V s =-W .(99) 

is the equation of shear for the cantilever Fig. n. 

If we lay off XB = — IF, the line (2) will be the line whose equa¬ 
tion is V s = — IF. This is called the line or curve oj shear. The 
shear is the same at every section. 

2. A Cantilever Uniformly Loaded.—If, in Fig. 11, the weight 
IF is removed, and the beam is one of uniform cross-section whose 
weight is considered, it becomes a uniformly loaded cantilever. 

Employing the same momenclature as before, and represent¬ 
ing by w the weight of the beam, per lineal inch, we have for 
the bending moment at any section, as c, 

'V* 

M = — wx • - ..(100) 

2 2 


This is the equation of bending moments. The line of bend¬ 
ing moments represented by this equation is a parabola, which 
may be constructed by points. Since M is negative, it will lie 
below the axis of A". If wl = W , the line of bending moments 
will be line (3) in Fig. n. 

As M is a numerically increasing function of it will have 
its greatest value when x = l. The dangerous section is therefore 
at X , and 



wl 2 

2 


(101 


By definition the shear at the section c is — wx, hence 

V s =-wx .(102) 

is the equation of shear. The line of shear is a right line. The 
abscissa of the point where it crosses the axis of # may be found 
by making V s = o and solving for and the point where it 
crosses the axis of V s bv making rr = o, and solving for F s ; if 







FLEXURE OR BENDING. 


59 


these points are the same, a second point may be determined 
by giving x some other value and solving for V s . 

If W = wlj line (4) in Fig. n will be the line of shear. As 
V s is a numerically increasing function of x, it will have its great¬ 
est value at X, where 

(V s ) m =-wl .(103) 


( 3 ) 


1 A 


Xu 


3. A Beam without Weight Resting on End Supports and 
Supporting a Load at the Middle Point.—Let OX, Fig. 12, 
represent the axis of the beam and let the nomenclature be as in 
the preceding case. 

The first step is to deter¬ 
mine the value of the reactions M - 

1 

Ri and R 2 . Since the load ! 

is at the middle point, they 0 -—-k- 

must be equal to each other, and «- iX 

, W R * L__ 

each equal to —-. 

2 Fig. 12. 

The bending moment at any 

section between O and A , taking the forces to the left of the sec¬ 
tion, is 

Wx , . 

M = — .(104) 


St) 


R., 


The line of bending moments between O and A is therefore 
a right line which passes through O, where x = o , at A, where 

1 . . . , Wl 

x=—, it has a positive ordinate of —. 

2 4 

The bending moment at any section between A and X, 
taking the forces from the left, is 


M = 


Wx 

2 



Wx Wl 

- + -. 

2 2 


(105) 


The line of bending moments between A and X is a right 
line which passes through X, where x=l\ at A, where x=- % 

it has a positive ordinate of • It is the line (1) in Tig. 12. 















6o 


CIVIL ENGINEERING. 


The dangerous section is evidently at A, and 

Wl 


M m = 


(i°6) 


W 

The shear at any section between O and A is +— , and 

2 

that at any point between A and X is 


Hence 


and 


2 2 


W 

V s = — from O to A } 
2 


W 

F. s = — — from A to X. 


(108) 


(109) 


The line of shear is therefore represented by line (2), Fig. 12. 
4. A Beam Resting on End Supports and Uniformly Loaded.—• 

Let OX, Fig. 12, be the axis of such a beam, and w be the uni¬ 
form load per inch of length. Let the nomenclature be as before. 

wl 

The reactions will then each be equal to —. 

The bending moment at any section between O and X will be 


, wlx wx 2 

M =-- 

2 2 


(no) 


This is the equation of a parabola, which passes through 
O, where x = o, and through X, where x=l. Its maximum 

l 

ordinate is at A, where x=z ~- The dangerous section is there¬ 
fore at A, and 



wl 2 

T" 


(in) 


If we make W = wl, the line of bending moments will be 
represented by line (3), Fig. 12. 










FLEXURE OR BENDING. 


61 


The shear at any section will be 

wl 

V s =—-wx .: (112) 

The line of shear is therefore a right line which passes through 

A, where x = ~, and crosses the axis of V s at a point — above O; 

2 

if W = wl, it is line (4), Fig. 12. 

The shear is greatest at the sections O and X , where 



and 


wl 

2 


(113) 


Since V s = 



we know from calculus that if 


M has a maxi¬ 


mum or minimum state, it will be found where -^=7 or V a = o. 

oX 

To determine whether it is a maximum or minimum state we must 

find the value of x from the equation ^47-=0, and substitute 

it in the second differential coefficient. If it makes the second 
differential coefficient negative, it is a maximum state, and if it 
makes the second differential coefficient positive, it is a minimum 
state. 

Since the dangerous section is at the section where the bending 
moment has its greatest numerical value, it is usually immaterial 
whether it is a maximum or minimum state of the ordinate of 
the curve of bending moments. 

It will be observed also that the section of mathematical 
maximum or minimum bending moment need not necessarily 
be the dangerous section, since the bending moment may be 
numerically greater at some other section of the beam. 

In plotting the curves of shear or bending moments approxi¬ 
mately, it should be remembered that a curve is convex upwards 
when the second differential coefficient of its ordinate with respect 
to x is negative, and concave, when the second differential coeffi¬ 
cient is positive. 

Modification of Cases 1 to 4.—If the cantilever in Case 1 has 
one or more additional forces applied between its extremity and 





62 


CIVIL ENGINEERING. 


the fixed end, the equations and lines of bending moments and 
shear may be deduced for each segment into which the beam is 
divided by the applied forces, by the application of the method 
above described. 

If the beam resting on end supports, in Case 3, has one or more 
forces applied between the middle point and supports, the "reac¬ 
tions will not usually be equal to each other. The intensities of 
the reactions may, however, always be determined by applying 
the equations of equilibrium for parallel vertical forces. 

Having determined the reactions, the equations and lines 
of bending moments and shear may be deduced by applying the 
methods above described. 

Application.—A beam AB, Fig. 13, 20 feet long, rests on two 
supports 16 feet apart, one of which is under the right end of 
the beam. The beam supports a uniform load of 20 pounds 



20 -_■>. 



c 


A. 


4' 3 ' 


X 0 



100 lbs. 

Fig. 13. 

per lineal foot and also a weight of 100 pounds, 13 feet from its 
right end. Deduce the equations of bending moments and 
shear. 

To Find the Reactions .—Let Ri = reaction at left support; 

R 2 = “ “ right “ 

Then 

-\-Ri +R 2 - ioo— (20X20) =0.( a) 


Since the resultant of the uniform load acts at the middle point of 
the beam, we have for the moments about B 


— (100X13)- (400X10)+ 7 ^iXi 6 = 
—1300 — 4000 + Ri X16 = 

Ri = 



53 °° 

16 ’ 

33 U- 












FLEXURE OR BENDING. 


63 


Substituting this value of Ri in equation ( a ) we have 

^2 = 5 °°- 33 i i = i 68|. 

Bending Moment. —Mark the left support C, and the point of 
application of the weight D , and assume the origin at A. 

2 ox 2 

From A to C M = — —— = — io.v 2 ; 

2 

“ C “ D ikf =—iox 2 +331 —4); 

“ D “ B M= — iox 2 + $i > i\(x— 4) — ioo(#—7). 

The curve of the bending moments in each segment of the 
beam is therefore a parabola. 

By substituting proper values for x in this equation, the bend¬ 
ing moment at any section of the beam may be readily ascertained, 
and by plotting the curves the position of the dangerous section 
can be found. 

Shear. 

From A to C V s = —20X] 

“ C “ D V 8 = — 20X+331J; 

“ D “ B V s = — 20.V + 331J —100. 

The line of shear in each segment is therefore a straight line. 
By substituting proper values for x, the shear at any section of the 
beam can be readily ascertained and the line of shear can be 
constructed. 

Dimensions of Beams of Uniform Section for Simple Load¬ 
ing.—If, in the equation 


we substitute the value of M m deduced for cantilevers and beams 
resting on end supports, we shall have 

p/r 

-— = IVI for a cantilever with a load W at its end, . . . (114) 

/ 

s l[L = L llA-“ “ “ ““ uniform load of w, . . (115) 

y' 2 




64 


CIVIL ENGINEERING. 


W7 

/ 


m 

4 


“ “ beam resting on end supports with a load W 


at its middle point, 


(n6) 


s"I wl 2 
/ = 8 


for a beam resting on end supports with a uniform 


load of w 


(117) 


In which s" = safe unit stress in pounds; 

-7 = section modulus; 

y 

W = weight in pounds; 

l = length of beam in inches; 

w = weight of uniform load per inch. 

By substituting for s" and the section modulus their proper 
values from the tables, we can determine any one of the five 
quantities b, d, /, W, and w, when all the others are known. 


PROBLEMS. 

24. A rectangular wooden cantilever 10 feet long and 6 inches 

deep must support a weight of 200 pounds at its outer extremity; 
what should be its width? The weight of the cantilever is not 
considered. Ans. 4 inches. 

25. A wooden beam 4 inches square, resting on end supports, 

must support a uniform load of 36 pounds per lineal foot, including 
its own weight; what is the maximum safe distance between the 
supports? Ans. 14 feet. 

26. What load may be safely hung from the middle point of 
a wooden beam of circular cross-section which rests on end supports 
10 feet apart? The radius of the beam is 4 inches and the weight 

7 d 3 

of the beam is not considered. -7= — . Ans. 1707 pounds. 

y' IO • • r 

27. A wooden beam, 10 feet long, the cross-section of which is 
an isosceles triangle whose horizontal base is 6 inches, rests on end 
supports. If it carries a uniform load, including its weight, of 120 
pounds per lineal foot, what must be the altitude of its cross- 

section? Ans. 8.48 inches. 

« 






FLEXURE OR BENDING. 65 

To Find the Strongest Rectangular Beam of Uniform Cross- 
section which can be Cut Out of a Circular Log. 

Let D = diameter of the log in inches; 
b = breadth of beam in inches; 
d = depth of beam in inches. 

The expression for the stress in the surface fiber at the danger¬ 
ous section of a beam is 


M m M„ 6 M m 
1 bd 2 bd 2 ' 
Y 6 


(118) 


The strongest beam is the one which has the least stress at 
the surface fiber at the dangerous section. It is therefore the 
beam in which the relation between b and d is governed by the 
condition that bd 2 in equation (118) shall be a maximum. 

Since the log is circular, 

b 2 + d 2 =D 2 , (119) 


From equation (119) we have 


d 2 =D 2 — tP .(120) 


Multiplying through by b we have 


bd 2 = bD 2 — b 3 .(121) 


Differentiating with respect to b we have 


d(bd 2 ) 

db 


= D 2 -$b 2 (122) 


and 


d 2 (bd 2 ) 
' (db ) 2 


= -6 b. 


(123) 


To determine the value of b which will make bd 2 a maxi- 












66 


CIVIL ENGINEERING. 


mum, we must place the first differential coefficient equal to 
zero and solve with respect to b : 


D 2 — T ) b 2 — o; 


(124) 


b = DV\ .(125 

This substitution in the second differential coefficient gives 
a negative value and therefore corresponds to a maximum. 

Substituting this value in equation (120) and solving for d 
we have 

J = Z)v / f.(126) 


To construct the cross-section, draw a circle with a diameter 
D. From either end of the diameter, lay off on it a distance 

equal to —, and at this point erect a perpendicular to the diam- 
3 

eter. From the point where the perpendicular intersects the 
circumference draw chords to the ends of the diameter. These 
chords will be the sides of the required inscribed rectangle. 

Dimensions of Beams of Uniform Strength.—If, in the equa- 

s"I 

tion — r~ = M i we substitute for M its values for the different 

/ 

beams above considered, we shall have 


s"I 

y 


- Wx, 


(127) 


s"I wx 2 

y ~ 2 1 

s"I Wx 

y 2 ’ 

s"I wlx wx 2 

y 2 2 


(128) 


(129) 


• • (13°) 


If, in equation (127), we substitute, from the table, for s" and 

















FLEXURE OR BENDING. 


6$ 

— their proper values, for a wooden beam of rectangular cross- 
section we shall have 


ioooX^f 2 

6 


= -Wx. 


(13 1 ) 


If we assume a value for b and W, and solve with respect 
to d 2 , we have 



6 Wx 
10006’ 


(132) 


in which d and # are the only variables. 

If we assume values of x } and deduce corresponding values 
of d , they will be the varying depths of cross-section of a wooden 

My' 

cantilever of uniform strength, since 


/ 


will be constant and 


equal to s" at every section of the beam. 

Equation (132) is the equation of a parabola. If the values 

of — are laid oh above and below the axis, the beam will be of 
2 

the form shown in Fig. 14, page 68. 

If we assume values for d and IE, and make b vary with x f 
the equation for a wooden cantilever loaded at the end, (127), 
becomes 


6 Wx_ 

ioood 2 


(i33) 


This is the equation of a right line, and the beam is of the 
form shown in Fig. 15, if the value of - is laid off on either side 
of the axis. 

The cross-section of the beams Figs. 14 and 15 must be 
enlarged near their free end, as shown in Fig. 15, to meet the 
requirement that the shearing strength of the sections near the 
end shall be sufficient to support the weight, or 


s 8 "bd = or > W. 









68 


CIYIL ENGINEERING. 


In the discussion above given we may substitute any other 
material for wood by substituting for s" its proper value; we 




may substitute any other form of cross-section for the rectangular, 

by substituting for ~ its proper value; and we may change 

the system of loading by substituting for M its proper value in 
terms of l and W or w. 


PROBLEMS. 

28. A wooden beam, resting on end supports 10 feet apart, 

must support a load of 1000 pounds at its middle point. The 

beam is of circular cross-section and of uniform strength to resist 

bending except at its end, where it is to be cylindrical in shape and 

of sufficient cross-section to resist shearing. Find diameter of 

cross-section at the end, at the middle point, and at the quarter 

I d 3 

points. — =—. 

y 10 

Ans. At end diameter 0.8 inch; at middle point 6.69 inches; at 
quarter points 5.31 inches. 

29. Determine the equation of the curve of meridian section of 
a beam of uniform strength which is of circular cross-section, rests 
on end supports, and is uniformly loaded. 

Ans. d 3 = (lx —x 2 )' 













CHAPTER IV. 


BEAMS FIXED AT THE ENDS, AND CURVE OF MEAN 

FIBER. 

A beam is fixed in position at any point by imposing the 
condition that the direction of its axis or its mean fiber at that 
point shall remain fixed. 

To determine the reactions of the supports, it is necessary to 
combine this equation of condition with the equations of equilib¬ 
rium already given. 



To deduce the equation of the mean fiber or the axis of a 
beam after deflection, let the cantilever shown in Fig. 16 be 
acted upon by the force F. 

Let AB = fixed section; 

CD = adj acent section; 

OH — ox = L = length of fiber ab before deflection; 
l = bc = elongation of fiber ab ; 
y= ordinate of fiber ab referred to neutral axis H\ 

E = coefficient of longitudinal elasticity; 
p = HE = radius of curvature of mean fiber after deflec¬ 
tion; 

5'"^ = unit stress in fiber ab, 

69 















70 


CIVIL ENGINEERING. 


Within the field of perfect elasticity we must then have 


El 

s'"y=z- 


(134) 


From similar triangles we have 

be: bH :: OH: OE, 

or l:y::dx:p; 

hence 

7 ydx 

i == • • • • 

p 


(13s) 


Substitute for dx its value L. 


,_yL l_y 

P L P 


(136) 


1. 

Substituting this value of — in equation (134) we have 


s'"y=—, or 

P P 


( I 37 ) 


From the theory of flexure or bending we have 


hence 


or 


M 

<.rrr _. 


M E 

1 P ’ • • • • 


EI 

• 

M =—. 

P 



dP 

P ~Sx 3 2 y’ .( I41 ) 


From calculus we have 












BEAMS FIXED AT THE ENDS, AND CURVE OF MEAN FIBER. 71 


in which dl = length of elementary part of a curve; 

3 x = its projection on the axis of X\ 

fiy _ < < <■' < < < < a ay. 

x and y — coordinates of points of the curve. 


If the deflection of the beam is very small, as it must be within 
the field of perfect elasticity, dl is sensibly equal to dx and we 
have 

Bx 2 

.( J 4 2 ) 


Substituting this value of p in equation (140) we have 


M = 


EId 2 y 
dx 2 ’ 


(M 3 ) 


in which y and # are coordinates of points of the curve. 

If, in this expression, the value of M can be expressed in 
terms of x, we can determine, by double integration, an expres¬ 
sion for y in terms of x, which will be the equation of the mean 
fiber. 

From this equation we can plot the curve of mean fiber and 
determine the maximum deflection of the axis of the beam under 
its load. 


Applications. 


I. Cantilever without Weight, Fixed Horizontally, Supporting 
a Load at its Extremity.—In Fig. 11, page 57, let 
OX = axis of beam before deflection 
= axis of X ; 

OM = axis of Y ; 

/ = length of beam; 
y w = maximum deflection of beam; 
c = any section of beam between O 
and X ; 

W = weight suspended from extrem¬ 
ity; 

M = bending moment at c. 









7 2 


CIVIL ENGINEERING. 


From the deduction, page 57, we have 


hence 



EId 2 y 

ox 2 



Integrating with respect to x we have 


Eldy 

ox 



(144) 


(145) 


oy 

in which is the tangent of the angle made by the mean fiber 

ooc 

with the axis of X, and C is the constant of integration. 

dy Wl 2 

By hypothesis, at X, where x = l, — = 0; hence C = ——• 

Substituting this value in (145) and integrating, 



Wx 3 Wl 2 x _ 


(146) 


where x = l, y— o; hence 


Wl 3 Wl 3 Wl 3 

C ~ 6 ~ 2 3 

W 

and y 6£/( lV3 3^ 2 x -f- 2/^). • • • • (147) 

Since y is a numerically decreasing function of x , it will have 
its greatest numerical value, within the limits of the beam, 
where ^ = 0 or 

Wl 3 

y,n ~ ~ 3 EI ■ 

% 

It will be observed that the expression for the bending moment 
at a.ny section is the second differential coefficient with respect 















BEAMS FIXED AT THE ENDS , AND CURVE OF MEAN FIBER. 73 

to x of the ordinate of curve of mean fiber at the same section, 
multiplied by the constant EL 

Therefore where the bending moment of a curve passes 
through zero, changing its sign, there will be a point of inflection 
in the curve of mean fiber. 

dM 

The shear at any section, , is the third differential coef- 

ox 

ficient of the ordinate of the mean fiber with respect to x, multi¬ 
plied by the same constant. 

The form of the curve of mean fiber is that approximately 
shown in Fig. 17; it is convex upwards, since for any value of 
x less than /, M, the second differential coefficient, is negative. 

II. Cantilever Fixed Horizontally and Loaded Uniformly.— 

Let w = load. in pounds per unit of length. 

From the deduction, page 58, Fig. n, we have 


Integrating, 


M = 


EId 2 y 

3 x 2 


wx 2 

• • 

2 


Eldy 

ox 


wx? „ 

~T +C ■ 


(148) 


(149) 


dy wl 3 

By hypothesis, where x = /, x^ = o; hence C = -^~ 

Integrating again, 



wx 4 
24 



wl 4 wl* wl 4 

where x =/, y = o; hence C' = — — -5” = ~ 


(15°) 


w 

and y=-—Ej(x*- 4 l 3 x+ 3 l -*). . . . (151) 

As in the preceding case, the curve is convex upwards, and 
of the form shown in Fig. 17. Its greatest ordinate is where 
x = o, or 

wl 4 

y ™ = ~&EI 


• (152) 

















74 


CIVIL ENGINEERING . 


III. Beam without Weight Resting on End Supports, with 
a Load W at Middle Point.—From previous deduction we have 
at c , Fig. 18, any section in left half of beam, 



Integrating, 


M = 


EId 2 y 

ox 2 


Wx 
' < 

2 


(153) 


Eldy Wx 2 
— 4 


+ C. 


(154) 


Because of symmetrical loading where x=—, 


o; hence 


C=- 


IF / 2 


16 

Substituting, 


Integrating, 


Eldy 

Wx 2 

TT 7 2 


dx 

4 

16 

(155) 


Wx 3 

Wl 2 x „ 


Ely = 

12 

16 +t • * * * 

(iso 


Where x = o, y — o; hence C 7 = o, and 

W 

y = 48 EI ~~ 3 ^ 2x ) • • • • ( I 57 ) 

Since M is positive for any value of x less than the curve is 

concave upwards and is of the form shown in Fig. 18. The 

maximum value of y is where x = —, or 

2 

WE 

y "~~4&EI .( x 5 8 ) 























BEAMS FIXED AT THE ENDS, AND CURVE OF MEAN FIBER. 75 

Since the beam is centrally loaded, the curve of mean fiber 
will be symmetrical with respect to a vertical line through its 
middle point. 

IV. Beam Resting on End Supports and Uniformly Loaded.— 

Let weight in pounds of load per unit of length. From 
previous deductions we have at C 


Integrating, 




EId 2 y 

dx 2 


wlx wx 2 

2 2 




El By wlx 2 7PX 3 

dx 4 6 


(159) 


(160) 


l Sy 

Because of symmetrical loading, where #=—, ^=0; ^ ence 
wP 


C = 


24 


. Substituting we have 


Integrating, 


Eldy 

wlx 2 

wxP 

wP 

dx 

4 

6 

24 

EIy= 

wlx 3 

wx 4 

wPx 

12 

24 

24 


. • ♦ * • (161) 


+C', • * . (162) 


where x = o y y = o; hence C' = o, and 


w 


y = ■' ( 2 ^ 3 —x 4 —Px) .(163) 


For any value of x less than /, M is positive; the curve of mean 
fiber is therefore concave upwards, and of the form shown in 

l 

Fig. 18. The value of y is a maximum where #=—, or 

57c/ 4 

^ m== ”384 El 


( 164 ) 
















7 6 


CIVIL ENGINEERING. 


V. A Beam without Weight Fixed Horizontally at Both 
Ends and Supporting a Weight W at its Middle Point.—In Fig. 19 



Fig. 19. 


let ^4 OX = axis of mean fiber of beam before deflection; 

0 = origin of coordinates; 

/ = length of beam, in inches, between the supports; 

W = a weight in pounds acting downwards at the middle 
point B of the beam; 

Ri=R 2 = reactions; 

V s ' = V ” = shears at R\ and R 2 ; 

c = any section between O and the middle point of the 
beam B\ 

Oc = x. 

The effect of the weight W alone is to deflect the mean fiber 
of the beam as shown in Fig. 18. To make its tangent at O, 
dy 

equal to zero, requires the introduction of an additional 

force F to the left of O, acting in the same direction as W. 
Let a = lever arm of F with respect to O. 


IV 

Ri=— + F=V/ +F, .(!65) 


and 



Integrating, 

Eldy IV x 2 


W~~~ Fax+c ' • • • 


-Fax + C .(167) 


By hypothesis, where x=o, ^=o; hence C=o, and 


Eldy Wx 2 


—Fax. 


dx 4 


. (168) 













BEAMS FIXED AT THE ENDS, AND CURVE OF MEAN FIBER . 77 


Because of symmetrical loading, where x=~, ~=o; hence 


2 ox 


Wl 2 Fal 


16 


-= o. 


• (169) 


and 


Fa = 


Wl 


8 


07 °) 


Substituting this value of Fa in equations (166) and (168) we 
have 


EId 2 y Wx Wl 
dx 2 ~ 2 ~ 8’ ‘ * e 


• • 071) 


Eldy Wx 2 Wlx 


dx 


8 


072) 


Integrating again, 


Ely 


Wx 3 _ Wlx 2 
12 16 


+c f , 


• • * 


073 ) 


where ^ = 0, y = o; hence C' = o, and 


W 

y = ^SEI (4X ' 3 “^ .(i74) 


The curve of mean fiber has a point of inflection where M — o, 


for which 


Wx Wl 


l 


— ~x~ = o, or #=—.(175) 


28' 4 


It is of the form shown in Fig. 19. 

The maximum deflection is at B, where the value of y is 


Wl 3 


tn 


192 El 


■ 076) 



















78 


CIVIL ENGINEERING. 


If the line of bending moments is constructed, it will be seen 

Wl . _ 

that the greatest values of M are + — at the middle point, and 

o 

Wl Wl 

at each end. Hence M w = ± 


8 - g • 

The equations of the line of shear are 


W W 

V 8 '= +— to the left of B , and V/= —— to the right of B. (177) 


VI. A Beam Fixed Horizontally at Both Ends and Uni¬ 
formly Loaded.—Assume the same notation as above and replace 
IF by a uniform load of w per lineal inch. Then at c 


M 


OX* 


1 wl \ wx 2 „, % wlx wx 2 

— + T )x - F(a+x )=— — —Fa 

2/2 2 2 


wx* 


— V/x - Fa 


(178) 


Integrating, 


Floy wlx 2 wx 3 


ox 


4 


— Fax+C. 


• • • • 


(179) 


dy 

By hypothesis, where x = o, v^ = o, hence C= o. 

Because of symmetrical loading, where x = ~> ^“=°; hence 


Fa = 


wl 2 

12 


(180) 


Substituting this value in equations (178) and (179) we have 

EId 2 y wlx wx 2 wl 2 


M= 


dx 2 


12 


} • • 


• (181) 


El dy wlx 2 wx 3 wl 2 x 


dx 


12 


. (182) 
























BEAMS FIXED AT THE ENDS, AND CURVE OF MEAN FIBER. 79 


Integrating, 



w'lx 3 m 4 
12 24 


wl 2 x 2 

24 


-I - C\ • 


where x = o, y = o; hence C' = o, and 


(183) 



w 

2+eI 


(2 lx 3 — X 4 — l 2 X 2 ). 


(184) 


The curve is of the form shown in Fig. 19, and has a point of 
inflection where M = o, or where 



The deflection is a maximum at the middle point where x=-: 

2 

hence 



The equation of bending moments 


,, wlx wx 2 wP 

M =--- 

2 212 


(187) 


is the equation of a parabola whose axis passes through B , and 


whose curve intersects the axis of X at points l*J — on either 

X 2 

wl 2 

side of the middle point, B. The greatest value of M is — 


12 


at the ends; at the middle point M = —. 

24 

If equations (166) and (178) are examined, it will be seen 
that the bending moment at any section, as c , is equal to the 
algebraic sum of three moments: the bending moment at O, the 
moment of the shear at O with respect to c, and the moment of 
the load between O and c with respect to c. 













8o 


CIVIL ENGINEERING. 


VII. Beam without Weight Fixed Horizontally at One End, 
Resting on a Support at the Other, and Supporting a Weight 
at the Middle Point.—In Fig. 20 let 



Fig. 20. 


0 = origin of coordinates; 

W = weight applied at middle 

point; 

R\= reaction at O = F+V/ 

Vs = R\ — F = shear at O; 

R 2 = reaction at right end = Vs "; 

Fa = moment required to keep neu¬ 
tral fiber at O horizontal. 


For c , any section between O and B, we have 


M = 


EId 2 y 

dx 2 


= -F(a+x) + (F + V/) x=-Fa+ V/x. 


(188) 


Integrating and 
C = o, we have 


remembering 


that where 


x = o, 



Eldy 

ox 


= —Fax + 


V/x 2 

• ••••• 

2 


hence 


(189) 


Integrating again and remembering that where x=o, y= o; hence 
C'=o, we have 



Fax 2 V/x 3 

2 * 


(190) 


For c', any section between B and R 2 we have 


M = =-F(a+x) + (F+V s ')x-W(x 


-9 


Wl 


- Fa + V/x -Wx-\ -. .... (191) 


Integrating, 


Eldy t 
dx 


= — Fax -f 


V/x 2 Wx 2 Wlx 


+ C. . (192) 


Integrating again, 





















BEAMS FIXED AT THE ENDS, AND CURVE OF MEAN FIBER. 81 


EIy= — 


Fax 2 VJx 3 Wx 3 TF/jc 2 


+ 


+ 


Cx-\-C '. . (193) 


where # = /, y = o; hence 


FaF VJF ( PF/ 3 PF/ 3 

O / “I - 7 Cl* 

2664 


Substituting, 


Fa* 2 F/* 3 PF* 3 PF/* 2 ^ T^/ 2 VJF 

FIy = — - -+-7—+- -+Cx + - 

2064 26 

WF WF 


+ 6 ~ 4 


Cl. (194) 


If the tangent to the mean fiber at B is represented by <J>, we 
shall have, from (189) and (192), 


FI tan </> = — 


Fal VJF 


8 ’ ’ * 


• • • (i95) 


_ , Fal VJF WF Wl 2 , ^ 

FI tan <56= •+—— —5-+ + C. . . (196) 

^ 2884 V ^ / 


Subtracting (195) from (196) we have 


C=- 


TF / 2 

8 * 


Representing the value of y, at B, by /, we have from (190) and 
(194), after substituting for C its value, 


FaF VJF 

£7/ =_ — + — 


(i97) 


Fly ' = - 


FaF F s 7 3 


8 1 48 


WF WF _WF 

48 + 16 16 


+■ 


FaF VJF WF 


WF WF 


2 


6 


6 


4 


(198) 



































82 


CIVIL ENGINEERING. 


Subtracting (197) from (198) we have 


FaP V S 'F Wl 3 

2 “6 “ “ 48 ’ 


or 


—Fa + 


V s 'l Wl 
3 ~ 2 4 * 


(199) 


The equation of equilibrium of moments about R2 is 


Wl 

Fa+V 8 'l—=o. 


(200) 


Subtracting (199) from (200) and dividing by /, 


2 V ^ T T T T 

s =—W, or F/=— W; .... (201) 


3 2 4 


hence 

and 


= ffclF 




• • • • 


. (202) 

• (203) 


Substituting these values in equation (189) we have 


- = - \wix + —Wx 2 . 
ox 16 32 


(204) 


If the point of maximum deflection is between O and B , at that 

dy 

point must reduce to o. Making this hypothesis and deducing 
the value of ^ we have 


*=tV;.(205) 

but this is beyond the middle point and equation (204) is not 
applicable. 

Substituting values of V/, etc., inequations (188), (189), (190), 













BEAMS FIXED AT THE ENDS, AND CURVE OF MEAN FIBER. 83 


(191), (192), (193), and (194), we have for the segment between 
B and 0 


EId s y 

dx 3 




(206) 


EId 2 y 

dx 2 


=M=-\wiWwx, . 

10 10 


(207) 


Eloy 3 11 

Wlx+—Wx 2 f 
ox 16 32 


(208) 


EIy= — &Wlx 2 + .. (209) 

and for the segment between B and R2 , 


UAf-=v s = l Aw-w=-Mv, . . . . 

dx 3 16 16 ’ 


UA!i = m=-\wx+\wi, . . . 

ox 2 16 16 


(210) 


. . (211) 


Eloy 5 S WP 

—^ = _ ELWx 2 + 4 Wlx - —, 
dx 32 16 8 


. (212) 


3 11 Wx 3 Wlx 2 

Ely = - —Wlx 2 + - -Wx 3 - —7- + — 
32 96 64 


1 2 Wl 3 11 Wl 3 

L. W Px + 4 — --wp + ILL _ 
8 16 2 90 0 


\m 3 +WP 

= -^Wx?+\*Wlx 2 -l s \Wl 2 x+izWl 3 . . (213) 

V 

From equations (207) and (211) it may be shown that the 
bending moment has its greatest value at O where 


M m ——Fa =— teWI. 


• . (214) 



















8 4 


CIVIL ENGINEERING. 


The deflection is greatest in the segment between B and O, 

oy 

at the point when — is equal to zero. 

ooc 

Under this hypothesis the maximum deflection will be found 
to be where x = J^/ approximately. 

If this value is substituted in equation (213), the maximum 
deflection will be found to be 


IT/ 3 

y ™ ~ 108 EI 


( 2I 5) 


VIII. Uniformly Loaded Beam Fixed Horizontally at One 
End and Resting on a Support at the Other. —Let the nomen¬ 
clature be the same as above, and let w be the weight per unit 
of length of the uniform load. Then 


M = 


EId 2 y 

dx 2 


= —Fa + Vs'x — 


wx- 


. (216) 


dy 

Integrating and remembering that where x = o, ^ =0, we have 


Eldy V 8 'x 2 wx 3 

Fax +——- 


ox 


6 ‘ 


• • • (217) 


Integrating and remembering that where x = o, y = o; we have 


EIy = 


Fax 2 VJx 3 wx 4 


24 


. (218) 


Where x = l, M in equation (216), and y in equation (218), 
are each equal to zero. Hence 


wl 2 

Fa+V/l -=0, 

2 


Fal 2 VJl 3 wl 4 


24 


2 


6 


= 0 . 















BEAMS FIXED AT THE ENDS, AND CURVE OF MEAN FIBER. 85 
Solving these equations for Vs and Fa we have 

F 8 ' = fw/, 


R 2 =V 8 ''=iwl, 
wl 2 


Fa = 


8 * 


Substituting these values we have 


EId*y 


OX'- 


=F S = -wx+^wl, . . . . 


(220) 


EId 2 y wx 2 5 wl 2 

8x 2 = ~~ + g wlx g"> 


. (22l) 


El By wx 3 c wl 2 x 

- =— <r + j6 wlx2 — 8 ~’ ■ • ■ ( 222 ) 


dx 


wx 4 c _ wl 2 x 2 

EIy= —-+ ~wlx 3 ———. . 

24 48 16 


• • (223) 


The bending moment will be greatest either at O or where 


F,=o. 


Wl 2 C Q 

At O, M=— -x-: where F s = o or x = ^l, M =— -wl 2 . 
m 8 8 128 


oy 0 

The greatest deflection will be where —=0, or where x=-—l 

OX 10 

approximately. 

Substituting this value of x in equation (223) we have 


wl 4 . . . 

y m = iRoEI a PP roximatel y.( 22 4) 















86 


CIVIL ENGINEERING. 


From data deduced in the preceding tables we may construct 
the following table: 

TABLE OF MAXIMUM VALUES OF BENDING MOMENTS, FIBER 
STRESS, AND DEFLECTIONS IN BEAMS OF UNIFORM 
CROSS-SECTION, WHOSE LENGTH IS /, AND WHOSE LOAD 
IS EITHER W OR u'l = W. 


Method of Loading and Supporting. 

Maximum 

Moments, 

-V m . 

Maximum 
Fiber Stress, 

s m ~ i • 

y 

Maximum 

Deflections, 

^m. 

Cantilever loaded at end. 

Wl 

Wl/ 

i 

1 Wl 3 

3 El 

Cantilever uniformly loaded. 

Wl 

Wl/ 

i Wl 3 

2 

2 / 

8 El 

Beam resting on end supports, loaded at 

Wl 

Wl/ 

i WP 

middle point. 

4 

4 / 

48 El 

Beam resting on end supports, uniformly 

Wl 

Wl/ 

5 I VP 

loaded. 

8 

8/ 

384 El 

Beam fixed horizontally at both ends, 

Wl 

Wl/ 

1 Wl 3 

loaded at middle. 

8 

8/ 

192 El 

Beam fixed horizontally at both ends, 

Wl 

Wl/ 

1 Wl 3 

uniformly loaded. 

12 

121 

384 El 

Beam fixed horizontally at one end and 




resting on end support at the other, 

3 Wl 

zWl/ 

1 Wl 3 

loaded at middle. 

16 

i 67 

108 El 

Beam fixed horizontally at one end and 

Wl 

Wl/ 

1 Wl 3 

resting on end support at the other, 

uniformly loaded. 

8 

8/ 

180 El 


Stiffness of Beams.—The stiffness of beams varies inversely 
with their maximum deflections under the same load. From the 
above table it follows, therefore, that the stiffness varies 
inversely with / 3 , the cube of the length; and directly with E, 
the coefficient of longitudinal elasticity of the material. In a 

bd 3 

rectangular beam I is equal to —, and the stiffness of beams 

varies directly with the first power oj the width and the third 
power oj the depth of the cross-section. 

Strength of Beams.—The strength of beams of the same 
material or their power to support additional loads varies in- 













































BEAMS FIXED AT THE ENDS, AND CURVE OF MEAN FIBER. 87 


versely with the stress on the surface fiber at the dangerous sec¬ 
tion. From the above table it follows, therefore, that the strength 
varies inversely with /, the length, and directly with the section 
modulus. If the beam is rectangular, the section modulus 



hence it varies directly with the first power oj the breadth 


and the second power oj the depth oj the cross-section. 

Maximum Deflections.—In the construction of the floors of 
buildings, to prevent injury to plastered ceilings, it is customary 

to limit the deflection in each beam to about of its span. 

Under this limitation the safe load of the beam may be deter¬ 
mined from the formula for maximum deflection, page 86, and 
not from that giving the maximum fiber stress, since the first may 
give a smaller safe load than the second. 

Resilience.—If a gradually applied load W is placed on the 
end of a cantilever without weight, the work performed by the 
weight, within the field of perfect elasticity, will be equal to the 
resilience of the beam, or 

Wy 

Resilience = work = ——, .... (225) 

2 


in which y m is the deflection of the end of the cantilever. 

Substituting for W its value in terms of 5, the stress at the 
dangerous section, from the equation 

Wl 

s = .(226) 

7 


or 



(227) 


and for y m , its value obtained by substituting the above value 
of W in the expression 


IF/ 3 


y 


m 



(228) 


sEIy'l sP 

y m = sl i3 ~ 3 £/’ • 


( 229 ) 












88 


CIVIL ENGINEERING. 


we have 


Resilience = work = 


si si 2 

yi 3 £ y 

2 


l_f_ I_ 

3 2 Ey'*' 


(230) 


Substituting for I its value Ar 2 , in which A is the area of 
cross-section and r is its radius of gyration with respect to the 
neutral axis, we have 


and 


1 s 2 r 2 

Resilience =- ^— 0 Al . . 

3 2£/- 


Resilience 1 s 2 r 2 

Tl ’ * 


(23 1 ) 


(232) 


for the resilience in a unit volume. 

5 2 

If the stress is increased until 5 = limit of elasticity, becomes 


the modulus of resilience of the material. 

Similar expressions may be determined for resilience of beams 
otherwise supported and loaded. 

Impact.—If a weight W falls through a distance h before it 
strikes the end of a cantilever without weight, the total amount 
of work performed before it ceases its motion downwards will be 


work = W (h+yj), .(233) 


in which IF = weight of W in pounds; 

h = distance of fall in inches; 

y m f = amount of deflection at end of cantilever in inches. 
If the stress is within the field of perfect elasticity, the work 
of the weight will be equal to the resilience, or 


W{h + y m ') 



( 234 ) 


in which IF' is the intensity of a static weight which applied at 
the end of the cantilever will produce the same stress and strain 
as the falling weight. 

If we substitute for IF' its value in terms of y w ', from the equa- 
1 IF'/ 3 t,EI 

tion y m ' = - or W' = —p~yj, we shall have 


7 FJ 

W(h + y m ') = ^ T y^, 


( 23 S) 









BEAMS FIXED AT THE ENDS , AND CURVE OF MEAN FIBER. 89 

from which we can determine y n ', when all the other quantities 
are given. 

If we substitute for ——— and yj their values from equations 
(229) and (230), we have 

TT7 / 7 si 2 \ 1 s 2 I 

W \ l + i>Ey r / 3 2 Ey ' 2 / * * * - ( 2 3 6 ) 

or 

W ( h+ sEy ,S ) = 3 2 Ey' 2 ^ 1, * * * * ( 2 37 ) 

from which we can determine the maximum stress at the danger¬ 
ous section when all the other quantities are given. 

Similar expressions can be determined for beams otherwise 
loaded and supported. 


PROBLEMS. 

30. What is the deflection produced by dropping a weight of 
100 pounds from a height of 10 inches on the middle point of a 
wooden beam 20 feet long, 4 inches wide, 12 inches deep, which 
rests on end supports? Weight of beam not considered. 

Ans. 0.85 inches. 

31. What is the greatest unit stress produced in the above 

problem? Ans. 1593.8 pounds. 

32. What is the safe uniformly distributed load of a 6-inch 
I beam resting on end supports 20 feet apart, if the deflection is 
limited to of the span. £=30,000,000; 1 = 24; weight of beam 
14.75 pounds per linear foot. Ans. 118.6 pounds per linear foot. 







/ 


. 

. 

CHAPTER V. 

CONTINUOUS BEAMS, HORIZONTAL SHEAR, COMBINED 
STRESSES, AND ECCENTRIC LOADING. 

Continuous Beams. —A continuous beam is one resting on 
more than two supports. 

From the definition of the bending moment it follows that 
an expression for the bending moment as a function of x can 
always be determined at every section of a beam, if we know 
its loads and the reactions at the supports. Having an expres¬ 
sion for the bending moments in terms of x, we can readily deduce 
the equations and construct the curves of mean fiber, bending 
moment, and vertical shear. 

If the loading is given, we can always determine the reactions 
at the supports, if the number of such reactions does not exceed 
the number of equations of equilibrium, which, for parallel 
forces, is only two. 

If the number of supports exceeds two, the reactions cannot 
therefore be determined by the use of the equations of equilib¬ 
rium alone; other equations are necessary. These are furnished 
by the theorem oj three moments , if the bending moments at the 
end supports are known or can be readily determined. The 
theorem applies therefore to beams not fixed at the ends. 

Theorem of Three Moments. —The deduction of the equation 
of the theorem of three moments depends upon the principles: 

i. That the bending moment at any support of a beam, as 
B , Fig. 21, is equal to the bending moment at the preceding sup¬ 
port A, plus the moment of the shear at the preceding support 
A about the point B, plus the moment of the load between the 
supports A and B about the point B. This was shown to be 
the case in the beam fixed horizontally at both ends, and in the 
beam fixed horizontally at one end and resting on an end 
support. 


90 



CONTINUOUS BEAMS, ETC. 


9 1 


2. Since the mean fiber of an unbroken beam must be con¬ 
tinuous, the tangent of the mean liber at B must be the same, 
whether we discuss the segment AB or the segment BC. 

To simplify the discussion, it will be assumed that each seg¬ 
ment of the beam is uniformly loaded. 

In Fig. 21 let A, B, C be three consecutive supports of the 
beam. 



<t>' 


<y 


w l 


B 


w'l' 


/\ 






R 2 

m 2 

Fig. 21. 




R 3 

m 3 


Let AB = l; 

BC = l'; 

w = load per lineal inch in AB) 
w'= “ “ “ “ “ BC ; 

M\, Mo, and M s = bending moments at A, B, and C; 

Ri, Ro, and R3 = reactions at A, B , and C; 

V/ f V a ", and F/" = vertical shears at sections adjacent to sup¬ 
ports A, B, and C on their right; 
cj)" = angle made by mean fiber of beam with axis 
of ^ at B. 


Taking the origin at A, and the forces from the left, we have 
for the bending moment at any section between A and B 


Integrating, 


EId 2 y _ _ . wx* 


M = » 9 ~ =M 1 + Vs x — 


ox 2 


Eldy V/x 2 wx s 

= Mix+ --7—+ C. . 


dx 


(238) 


(239) 


7 °y 1 

For x = l, — = tan <p 


n 


hence 


C = EI tan 4 >”-Mil- 


V/l 2 



2 


. . (240) 













92 


CIVIL ENGINEERING . 


Substituting this value of C in equation (239), 

fdy \ V s 'x 2 wx 3 V s 'l 2 wl 3 

El(/ x -t m r)=M 1 x+- -+ _. . ( 24 I> 


Integrating, 

El (y — tan ft'x) 

M \X 2 V s 'x 3 wx 4 


+ 


V Jl 2 x wl 3 x 


24 


- M ilx - —— + — + C’, (242) 


where x=o, y — o, hence C' =0. Making #=/, y = o, and we 
have 

_ Mil V'P t ^/3 ? F s '/ 2 wl 3 , x 

— El tan <£" =-+ —7-——-—Mi/ — --+ -7- (243) 

^2 6 24 2 6 v 0 


Mi/ F// 2 , wl 3 

— + 


8 * * 


• • 


(244) 


Making x — l and M = M 2 in equation (238) we have 

wl 2 

M 2 =Mi + V/l -, . 


• • 


(245) 


hence 


v >JhJh™L 

8 1 1 + 2 ■ 


. (246) 


Substituting this value in equation (244) we have 


— El tan </>" = — 


Mol Mil wP 

3 6 24 


• • (247) 


Taking the origin at B, we have for the bending moment 
at any section between B and C 


w'x 2 


M=EI^=M 2 + V,"x 


• (248) 


5 k 

oy 


Integrating and remembering that where #=0, -^- = tan <£", we 
have 
























CONTINUOUS BEAMS , ETC. 


93 


(dy_ 


E\f x -^4>")=M 2 x+ 2 


V/'x 2 w'x 3 


• • 


Integrating and remembering that where x=o, y— o, 
C'=o, 

T?T/ M 2 x 2 V/'x 3 w'x 4 

El (y — tan d>"#) =-+—7—-. . . 

2 6 24 

Making x = l', where y = o, we have 


... M 2 V , V/'l ' 2 w'V* 

— El tan d) =-+ —7— —-. 

2 0 24 


Making x = l' and M = M 3 in equation (248) we have 


w'l ' 2 M 3 M 2 w'l' 

M 3 = M 2 + V/'l' -, hence V." - 

2 L i 2 


( 2 49 ) 

hence 

( 2 5°) 

(251) 

(252) 


Substituting this value of V s " in equation (251) we have 


-El tan cj)" 


M 2 l' 

2 


M 3 l' 

6 


M 2 l' w'l ' 3 
6 ^"12 


M 2 l' M 3 l' w'l ' 3 

-+ —7 — +- 

3 6 24 


w'l ' 3 

24 


(253) 


Equating the two values of —El tan <j> in equations (247) and 
(253) and transposing we have 

wl 3 w'l ' 3 

Mil-\- 2M 2 (l-\- 1 ') VM 3 l' - -— . . . (254) 

4 4 

This is the equation of the theorem 0 / three moments deduced 
for continuous beams whose segments are unequal but each 
uniformly loaded. 

It is evident that we can obtain as many such equations as 
there are supports to the beam less two. Thus we can deter¬ 
mine (n — 2) independent equations for a beam resting on n 

























94 


CIVIL ENGINEERING. 


supports. If, therefore, we know the bending moments at two 
of the supports, we can determine all the others. 

If the ends of a continuous beam are not fixed, the bend¬ 
ing moments at these supports can always be determined when 
we know the forces acting on the beam outside of these sup¬ 
ports. By substituting these values for M\ and M n in the 
equations above given, the bending moments at all the other 
supports may be ascertained. 

Since the bending moment at each support is simply the 
resultant moment of the reactions and weights on one side of 
the support, we can form enough equations of condition con¬ 
taining reactions, loads, and bending moments at the supports 
to determine all the reactions. 

Applications. 

I. Beam Uniformly Loaded Resting on Three Supports 
Equally Spaced. 

wl 3 w'l ' 3 

M\l-\- 2 M 2 (l~\~l r ) -\-M 2 , 1 — ~ ~7j~— "• . . (255) 

If 1 = 1 ' and w = w', this becomes 

, wl 3 

M\l-\- qAf 2 l + M 3/ =— . .... (256) 

2 


If the beam terminates at the end supports, by definition, 
M 1=0 and M 3 = o, and 


4M 2 l = — 


wl 3 


2 


) 


M 2 


wl 2 

IP 


If R\, R2 f and R3 are the total reactions at the supports, we have 

wl 2 wl 2 

(a) M 2 = -~g~ = Ril — —, .'. = .( 2 57 ) 

(b ) M 3 = 2R1I + R 2 l — 2 wl 2 = f7 vl 2 + R 2 l — 2 wl 2 = o, 

.(258) 

/ 

(c) R l + R 2 + R 3 = 2wl = %wl + ^wl + R 3 , R 3 = lwl. (259) 









CONTINUOUS BEAMS , ETC. 


95 


The expression for the bending moment in the segment AB , 
Fig. 21, is therefore 


,, _ WX 2 2 , WX 2 

M = RiX — -=7T wlx -. 

2 6 2 


. . . (260) 


In the segment BC , taking the origin at B, 


IV X 2 T C WX 2 

M=M 2 + v/'x -- - ^wl 2 + ±wlx -. . (261) 

208 2 


In equation (261) V" is the shear at B. 

From these equations we may determine the equations and 
curves of bending moments, vertical shear, and mean fiber. 

II. Beam Uniformly Loaded Resting on Four Supports 
Equally Spaced. 

Mil +2 M 2 (l +/' )+MJ' = -\wP-\w'l'* y . . (262) 

M 2 V + 2 M 3 (/' + /") + MJ," = - \v/V 3 - . (263) 

If l = = w=w' =w ", and M 1 =Ma=o, we have 

4M2/ + A/ 3/ = — .(264) 

M 2 / + 4M 3/ = — .(265) 

Combining and solving we have 

M 2 = il / 3 = - jqwI 2 . 

If Ri, fl 2 . ^3> and are the total reactions at the supports, we 
have 

wl 2 

M 2 =Ril -— = - h wl > 

2 




( 266 ) 








9 6 


CIVIL ENGINEERING. 


M 3 = 2RiI + R 2 l - 2 wl 2 = f- 0 wl 2 + R 2 l- 2 wl 2 = - tVw/ 2 , 

R 2 = ... ... (267) 

if 4 = 3IW + 2^2 / + ^3/ - f^ 2 = if w/ 2 + ff wl 2 + £3/ - \wl 2 = o, 


R 3 = HwL 


(268) 


i?l + i^2 + -^3 + -^4 = = T 4 0^ + i f ^ + rf' + ^4) 


.*. R 4 = i %w! 


(269) 


The bending moments in these sections are 


4 UUX 2 UJX 2 

M = y^wIx — ~^—, origin at first support = i?i# ——. . (270) 


_, I JO 5 7 wov 

M = —— wl 2 j r — WlX — - 

IO IO 2 


origin at second support = M 2+ V 8 "x ■ 


wx l 


; • (271) 


__ i o wr 

M = — — wl 2 + —wlx - 

IO IO 2 


woe 4 


origin at third support =M 3 + V s "'x — -. . (272) 

2 


In which V 8 " and V s "' are the shears at the second and third 
supports. 

III. Beam Uniformly Loaded Resting on Five Supports 
Equally Spaced. 

wP w'l' 3 

M l l+2M 2 (l+l')+M 3 l' ---. . . (273) 

4 4 


w'l' s w"l " 3 

M 2 l' + 2 M 3 (l' + l'') + M 4 I''=-—-^-, . . f 274 ) 















CONTINUOUS BEAMS , ETC. 


97 


W/////3 no/"/'" 3 

If 3 /" + 2M 4 (/"+/'")+M 5/"' = — —-j— — ——-—. . (275) 

If Mi = ilf5=0, w = w r =w" = w"' and we have 


+ 4M 2 1 + if 3/ = — 


w / 3 


(276) 


TM 2/+4M 3/ + M 4/ = — 


wl 3 


• * ( 2 77 ) 


wl 3 

+M 3 /+4M 4 /= ——. ...... (278) 

2 


Solving these equations we have 

M 2 = M 4 = -2 3 sWl 2 , M 3 = -^wl 2 . . . . (279) 

If the total reactions at the supports are Ri, R 2 , i?3, i? 4 , and R 5 , 
we have 


3 wl 2 1 1 

M 2 = — -^wl 2 =Ril -—, .*. Ri=-^wl .(280) 


28 


M 3 = — Js^l 2 = 2R1I + 7 ^ — 2 wl 2 = %\wl 2 + R 2 l — 2 wl 2 , 

.*. R 2 = ^wl .(281) 

We may write out the other equations or assume from sym¬ 
metry that 

R 5 = Ri = ^wl, R± = R 2 = i%wl, .... (282) 


and 


R 3 = ^wl — (R\-\- R 2 -\- R^-\- R^) — j^wl. . . . (283) 


The expressions for the moments will be 


1 t wx 2 wx 2 

M=—wlx — - = R\X— -, 

28 2 2 


(284) 


7 m 1 c; woe* 

M = - -^wl 2 +~wlx - — 
28 28 2 


wx- 


M 2 + V/'x-—,. (285) 

















9 8 


CIVIL ENGINEERING. 


M=-^cwl 2 + if wlx -— = Ms + V,'"x - —,. ( 2 86) 

28 25 2 2 V ' 

M = -\wl 2 +^\wlx-~ = ilf 4 + F 8 iv x - —. . (287) 

28 28 2 2 


From these expressions we may determine the equations and 
curves of bending moments, vertical shear, and mean fiber. 

Horizontal Shear in Beams of Uniform Cross-section.—The 
horizontal shear in beams is the shear in surfaces parallel to the 
neutral surface; the movement of these surfaces on each other 
may be readily illustrated by making a beam of thin superposed 
strips. 

In the cantilever fixed at its left end, shown in Fig. 22, let 



OX = 
OY = 
OZ = 
AABB = 
/ = 

CCDD = 
l T ox — 
W = 
M = 
M + dM = 
oo f o"o = 

00' = 
o n o ,n = 


its axis = the axis of X ; 
axis of Y ; 
axis of Z ; 

plane of cross-section; 

distance in inches of plane A ABB from free end of 
beam; 

plane of cross-section; 

distance in inches of plane CCDD from end of beam; 
a concentrated weight applied at free end; 

Wl = bending moment at section A ABB) 

W(l + dx)= bending moment at section CCDD ; 
neutral surface of beam between the two planes of 
cross-section; 

neutral axis of plane A ABB) 
neutral axis of plane CCDD) 
























CONTINUOUS BEAMS, ETC. 


99 


y — distance of any fiber of beam parallel to the neutral 
surface from that surface; 
y" = distance of plane aacc from neutral surface; 
b = width of beam; 
d = depth of beam; 

5 i /// = unit bending stress at a unit’s distance from oo r in 
the plane A B\ 

S2 /,,== unit bending stress at a unit’s distance from o"o' rf 
in the plane CD ; 

s s = unit shearing stress in the horizontal plane aacc . 

Then $i'"^ = unit tensile stress in a fiber of the plane AB at & 

distance y from oo r ; , 

$2 ff y ~unit tensile stress in a fiber of the plane CD at &• 
distance y from o n o ,n ; 

S\ n ydyoz = stress in the elementary area dydz of the plane AB 
at a distance y from the neutral axis; 

S2 ff ydydz = stress in the elementary area dydz of the plane CD 
at a distance y from the neutral axis. 



ydyoz = tensile stress on area AAaa of the 
plane AB, . . . * . . (288) 


yoydz — tensile stress on area CCcc of the 
plane CD .(289) 


In the above equations, each double integral is the sum of 
the products of all the elementary areas of the total area con¬ 
sidered, by their distances from the neutral axis; hence each 
is the static moment of the area considered about the neutral 

axis. 

If A' = area AAaa of the plane AB or of CCcc of the plane CD, 
/ = distance of center of gravity of AAaa from 00' or of 
CCcc from o"o"', 

then 

A'y' = static moment of area AAaa with respect to 00', or 
of area CCcc with respect to o'V". 

Substituting this value for the double integral in equations 
(288) and (289) we have 


L.of C. 



loo 


CIVIL ENGINEERING. 


Si"'A'y' = stress in area AAaa, 
s 2 '"A'y' = stress in area CCcc, and 
(s 2 " — S\")A’y' = difference of tensile stress on areas AAaa and 

CCcc. 

Since the small prism AAaaCCcc is free on all sides but aacc, 
and is in a state of rest, the shearing stress on the base aacc must 
be equal to the difference between the tensile stresses on AAaa 
and CCcc, or 

s 8 bdx={s 2 " — Si"')A'y' .(290) 


From equation 82 we have 


~ /// _ 

Si — 


M 


I 


and s 2 


/// 


M + dM 
I 3 


(251) 


in which / = moment of inertia of cross-section A ABB or CCDD 
about the neutral axis. 

Substituting in equation (29c) we have 


Sobdx = 


f VM + dMl 

Ml 

1 L / J 

“/ 1 


(dM)A'/ 

I ' 5 


(292) 


or 




f dM\ A'y' 
ox ) bl 


( 293 ) 


Mi Tr 
But ~N~ = V S 
ox 


= vertical shear in the cross-section A ABB. 


Sub¬ 


stituting this value in equation (293) we have 

V s A'y' 

Ss ~ bi ’ • • 


(294) 


or, the unit horizontal stress, at any point of a surface parallel to 
ike neutral surface, is equal to the quotient obtained by dividing the 
product of the vertical shear at the point, and the static moment of 













CONTINUOUS BEAMS , ETC. 


IOI 


the area of cross-section between this surface and the top of the 
beam , by the product of the breadth of the beam and the moment of 
inertia of the cross-section of the beam at the point considered. The 
units employed are inches and pounds. 

If H s = total shear in any horizontal surface of a beam, 

(H s ) m = maximum value of H s , 

s s " = allowable unit stress in shear, 

A = area of horizontal surface in shear, 

for safety s 8 "A =or > 

If the beam is of uniform cross-section, b and / are con¬ 
stant. Hence in the same horizontal surface, for which A'y' is 
constant, the value of will vary with V s ; or, the unit shearing 
stress in any surface of a beam of uniform cross-section , which is 
parallel to the neutral surface , will be uniform along the line cut 
out by any plane of cross-section , and will vary with the vertical 
shear along any line parallel to the axis of the beam. 

The unit horizontal shear will therefore be uniform at every 
point of a horizontal surface of a beam whose line of shear is 
parallel to the axis of X, and the total shear will be equal to 
s s bl. The unit horizontal shear will be variable in the horizontal 
surfaces of a beam whose line of shear is not parallel to the axis of 
X. The total shear in the plane will be equal to the mean value 
of multiplied by bl. 

In the same plane of cross-section V s will be constant, and the 
value of will vary with A'y'; or, the unit horizontal shear in any 
plane of cross-section will be uniform along any line parallel to the 
neutral axis , and will vary with the static moment of the area 
included between the surface of the beam and the surface of hori¬ 
zontal shear , along any line perpendicular to the neutral axis. 

In a rectangular beam of uniform cross-section whose depth 
is d and whose breadth is 6, 



d 

-+y 

2 y 


and 



The unit stress will, in any plane of cross-section, vary in a 




102 


CIVIL ENGINEERING. 


direction perpendicular to the neutral axis with the expression 

y 2 ; it will be zero at the surface where y = —, and will be a 

2 

maximum at the neutral surface where y = zero. The expression 



_ V,A'f _ V. b 

bl bl' 2 



(296) 


is the equation of a parabola when and y are the only vari¬ 
ables; hence the unit horizontal shearing stress in a cross-section 
of a rectangular beam will, in the direction perpendicular to 
the neutral axis, vary with the ordinates of a parabola. 

In Chapter II we have shown that the unit vertical and 
horizontal shears at any point in a beam must be equal to each 
other, hence the vertical shear cannot be uniform in the plane 
of cross-section as heretofore assumed, but must also be a maxi¬ 
mum at the neutral axis. 

The error introduced in assuming that the unit vertical shear 
is uniform in the cross-section of a beam is unimportant, since 
beams are designed ordinarily to resist bending forces and have 
a large excess strength to resist vertical shear. 

If, in equation (296), we substitute for / its value jjbd 3 , and 
for y its value zero at the neutral axis, we have 

*“K©’. (297) 

or the unit horizontal shear at the neutral fiber is f the unit verti¬ 
cal shear assumed to be uniformly distributed over the area of 
cross-section. 

In a rectangular beam, therefore, the true unit vertical shear 
at the axis is f its assumed value. 

PROBLEMS. 

33. A cantilever 6 inches wide, 8 inches deep, 10 feet long, 
supports a weight of 1000 pounds at its outer extremity. Find 
the maximum unit horizontal shear in the beam. Ans. 3if pounds. 

34. A wooden built-up beam 6 inches wide and 12 inches deep 
is made of three 4 X 6-inch beams bolted together. When the 
beam supports a load of 2000 pounds at its middle point, what is 






CONTINUOUS BEAMS, ETC. 103 

the unit shear at the planes of contact, and what is the total shear 
on the bolts? Ans. Unit shear 18 pounds. 

Total “ 12,960 

35. The 8-inch 18-pound I beam is designed to support a load 
of 5000 pounds placed at its middle point when resting on supports 
15 feet apart. Determine the unit horizontal shear at the neutral 
axis, assuming that the center of gravity of each half-section is 
3.6 inches from the neutral axis.* Ans . 1564 pounds. 

Combined Stresses. 

Tension and Flexure.—If a beam is subjected simultaneously 
to tensile and bending forces, the maximum unit stress will be 
in the fiber in which the sum of the tensile stresses due to the 
two forces will be a maximum. This fiber will ordinarily be 
the surface fiber on the tension side of the dangerous section. 

Let F = intensity of the tensile force; 

M = bending moment at dangerous section due to the 
bending force; 

A = area of cross-section of beam; 

I = moment of inertia of A about the neutral axis; 
y = distance from the extreme fiber to the neutral axis; 
s m = maximum longitudinal unit fiber stress in beam. Then 

F My ' 

Sm = A + I " 

For safety s m should not exceed the 
allowable stress of the material either in 
tension or in flexure. 

Fig. 23 represents an inclined rect¬ 
angular beam of uniform cross-section, 
one end of which is fastened in a wall. 

At the free end it bears a weight W, 
whose action line makes an angle with 
the axis of the beam. Its component 
W cos <j) produces tension in the beam, 
and its component W sin 0 produces 
flexure. 


• ••••• (,298) 



* Cambria Handbook, p. 156. 









104 


CIVIL ENGINEERING . 


Let l =length of the beam from the free end to the dangerous 
section; 

b =;breadth of cross-section; 
d = depth of cross-section. 

Substituting the above value in equation (298) we have 

W cos <f> 6 Wl sin (f> 

Sm= bd + bd 2 ’ • • • • ( 2 99 ) 

in which s tn is the stress in the fibers of the upper surface in the 
section at the wall. 

Compression and Flexure.—If a beam is subjected simul¬ 
taneously to compressive and bending forces, the maximum 
unit stress will be in the fiber in which the sum of the compres¬ 
sive stresses due to the two forces is a maximum. This fiber 
will ordinarily be the surface fiber on the compression side of 
the dangerous section. 

For compression and bending equation (298) will give the 
maximum longitudinal unit fibre stress if F = the intensity of 
the compressive force. 

For safety s m should not exceed the allowable stress of the 
material either in compression or flexure. 

If in Fig. 23 the beam is inclined upwards instead of down¬ 
wards, making the same angle with the vertical, equation (299) 
will give the stresses in the fibers of the lower surface in the 
section at the wall. 


PROBLEM. 

36. A square wooden cantilever inclined at an angle of 60 degrees 
projects 12 feet from a wall. What should be its dimensions to 
sustain a weight of 300 pounds if the allowable unit stress is 1000 
pounds ? Ans. 6.09 inches square. 

Flexure and Shear.—A force of flexure produces longitudinal 
stresses in the fibers of the cross-section of a beam, and shearing 
stresses in each horizontal plane and in each vertical plane of 
cross-section. 





CONTINUOUS BEAMS, ETC. 


!°5 

Thus if ABCD , Fig. 24, represents an elementary particle 
in the tensile side of the beam, on the 
face AC there will be a bending and shear¬ 
ing stress, and on the face AB a shearing 
stress. 

The joint effect of the stresses will be 
to bring a combined stress on some oblique 
plane BC. ^ 

Let it be required to find the maximum 
unit bending stress and the maximum unit shear on the plane 
BC. 

Let s = unit bending stress on AC; 

s s = “ shear on AC and AB. Then 

5XTC = total bending stress on AC; 
s s xAC= “ shear on AC; 
s s xAB= “ “ “ AB. 



Let 0 = angle ABC; 

5' = unit bending stress on BC; 
s s ' = “ shear on BC. 


Resolving the three elementary stresses into their compo¬ 
nents, perpendicular and parallel to BC, we have for the total 
bending stress on BC 


s' XBC = sXAC sin cj> + s s xAC cos cp + s s xAB sin <£, . (300) 


or 


s' = 


xX AC sin <p cos cp s s AB sin cp 


BC 


BC 


BC 


. . . . (301) 


But 


AC . , , AB 

— = sm cp and ^r = c°s cp; 


nence 


5' = 5 sin 2 cp + 2S s sin cp cos cp. 








io6 


CIVIL ENGINEERING. 


i 


Substituting for sin <f> its v 
its value sin 2 (/>, 



I — COS 2 (f) 

2 


} 


and for 2 sin cf> cos (f> 


s' = \s — Js cos 2<j)-\-s s sin 2 (f> .(3 02 ) 

To determine the value of 2<j> which will make s' a maximum, 
we have 

sin 2(p + s s cos 2(p .(303) 

0(29) 2 

Placing the first differential coefficient equal to zero, 


tan 2(f) = — 


2Ss 

s 


(304) 


There may then be two angles 2 (f>, differing by 180 degrees, 
one with a positive sine and negative cosine, the other with a neg¬ 
ative sine and positive cosine, each giving a mathematical maxi¬ 
mum or minimum, the numerically greater of which will be the 
value sought. That is, 


sin 2(f>= ± 

COS 2 (f) = T 


V s 2 + 4S s 2 ’ 
s 

\A 2 + 4s 2 


r. 


( 305 ) 


By substitution in (302) we have 




( 3 ° 6 ) 


Obviously the upper sign will give the greater, and hence the 
greatest, numerical value of unit bending stress on any oblique 
plane due to combined bending and shearing stresses. 

For the total shear on BC we have 


s 8 'XBC = sXAC cos (f> + s s xAB cos (f> — s s X AC sin <£, (307) 












CONTINUOUS BEAMS. ETC. 


107 


, AC AB AC . , 

° r 55 =S BC C0S ^ + BC cos cj)-s s X^ sin 0 . . . (308) 

= s sin 0 cos 9S + 5 s cos 2 <p — s s sin 2 0 

—s sin 0 cos 0 + s 5 (cos 2 0 —sin 2 0). 

sin 20 

Substituting for sin 0 cos 0 its value-, and for cos 2 0 — sin 2 0 

2 

its value cos 20, we have 

s/ = sin 2 (f) + s s cos 20.(309) 

Differentiating with respect to 20, we have 

OSs 1 

V/ '\ = ~ s cos 2d> — sin 26 .(310) 

0(29) 2 vo 

Placing the first differential coefficient equal to zero, and solving, 
we have 


Hence 


tan 20 = 


5 

2S s 


(3 11 ) 


COS 20= ± 


2 S, 


V j 2 + 4^ 2 ’ 


sin 20= ± 


Y ^ 2 + 4^ s 2 


( 312 ) 


Substituting these values in equation (309), the maximum value 
oj the unit shearing stress on any oblique plane due to a combined 
bending and shearing stress becomes 


0 /) w = ± 



( 3 J 3 ) 


In equations (304), (305) and (306), if s s = o, as it is in the 
surface fiber , then tan 20 = o, cos 20= — 1, sin 20 = o, 20 = 180°, 
0 = go°, or BC and AC coincide and (s') m = (s) m , as it should 















ic8 


CIVIL ENGINEERING. 


under this hypothesis. If s = o, as it is at the neutral fiber , then 
tan 20= oo, cos 20 = o, sin 20 = i, 20 = 90°, 0 = 45°, and {s') m = s s . 
In equations (311), (312) and (313) if s s = o, tan20=oo, 

s 

cos 20 = 0, sin 20 = 1, 20 = 90°, 0 = 45°, (V) w = - , which is the 

conclusion deduced on page 35; and if 5 = 0, tan 20 = 0, cos 20= 1, 
sin 20 = o, 20 = 360°, 0 = i8o°. Then (s 8 ') m = s 8 , which is the case 
in simple shear. 

For safety the maximum value of s' for the surface fiber of 
the beam at the dangerous section must not exceed s", the safe 
unit stress in bending for the material, nor must s/ at the neutral 
fiber in the section of greatest bending moment exceed s 8 ", the 
safe unit stress in shear. The value of s at any distance from the 
neutral fiber may be found from equation (87), 


5 = 


s'"y 


My 

I * 


The value of may be found from equation (294), 

Vs Ay 
s s bI . 

All quantities in these equations must be expressed in inches 
and pounds. 


PROBLEM. 

37. In a cantilever 6 inches wide, 8 inches deep, 10 feet long, 
carrying a weight of 1000 pounds at its outer extremity, find the 
maximum tensile and shearing stress in surface fiber, in neutral 
fiber, and in fiber midway between surface and axis at the dangerous 
section. 

Ans. (s')m =i8 75 pounds at surface, o pounds at neutral fiber, 
937.9 pounds midway between surface and neutral fiber. 

o pounds at surface, 31.25 pounds at neutral 
fiber, 469.2 pounds midway between surface and neutral fiber. 

Longitudinal Stress and Torsion.— As torsion is a shearing 

O 

stress, the formulas above given are also applied to stresses pro¬ 
duced by a combined longitudinal and torsional force. 

Let 5 = unit longitudinal stress in the fiber considered; 
s t = “ torsional ‘ ‘ “ “ ‘ ‘ “ 





CONTINUOUS BEAMS, ETC. 109 

5'= maximum unit longitudinal stress in fiber considered 
due to combined stress; 

s/ = maximum unit shearing stress in fiber considered due 
to combined stress. Then 




( 3 ! 4 ) 



• • • • • • * (3 1 5) 


These formulas may also be applied to the determination of 
the combined stresses of bending and torsion in fibers farthest 
from the neutral axis of a circular shaft subjected to bending and 
torsion. They may therefore be applied to the determination of 
the combined stresses in the surface fiber of the dangerous sec¬ 
tion due to flexure. This follows from the fact that in these fibers 
there is no shear due to the bending force. 

The value of s may be obtained from equation (88), 



and s t from equation (66), 


MR 

S t J * 

* P 

All quantities in these equations must be expressed in inches 
and pounds. 

PROBLEMS. 

38. A circular steel shaft 6 inches in diameter, resting on sup¬ 
ports 10 feet apart, transmits 50 horse-power when making 225 
revolutions per minute. Find the maximum unit tensile and 
unit torsional stress in the surface fibers in a vertical plane through 
the axis due to the combined bending and torsional moments. 

Ans. 815.5 pounds per square inch. 

39. A circular wrought-iron shaft 12 feet long between bearings 
while making 130 revolutions per minute transmits 50 horse¬ 
power. What should be its diameter if the pulleys are so placed 






no 


Cl 1 / 1 L ENGINEERING. 


as to produce a bending moment of 600 (foot-pounds) at its middle 
point and the maximum combined stress in the upper fiber is 
limited to 10,000 pounds? Ans. 2.54 inches. 

Distribution of Stress under Eccentric Loading. —The cross- 
section of a rod or column is said to be centrally loaded when the 
action line of the resultant force of tension or compression parallel 
to its axis coincides with that axis. It is said to be eccentrically 
loaded when this action line does not coincide with the axis. 

The theoretical distribution of stress over the area of cross- 
section depends on the hypotheses: 

1. The plane of cross-section does not warp or change form 
under the action of the force. 

2. All fibers offer equal resistances to tensile and compressive 
forces. 

3. The amount of stress developed in a fiber is proportional 
to its strain. 

Let ABCD, Fig. 25, be any cross-section of a column whose 





lower end is firmly supported, and let EK, Fig. 25, be the line cut 
out of this cross-section by a vertical plane. 

Let 0 = center of gravity of A BCD; 

A =area of cross-section; 

EK and GH = axes of symmetry; 

d b 

EK = d ; IM =---; /£ = -; 

3 3 

GH = b ; 

F = resultant force of compression, parallel to the axis 
of the column and perpendicular to ABCD ; 



































































































































































CONTINUOUS BEAMS, ETC. 


hi 


o' = center of pressure or point of application of F in 
the plane A BCD) 

s = unit longitudinal stress, due to F, at any point in 
the area A BCD) 
o = origin of coordinates; 
oK = axis of X ; 
oG = axis of Y ; 

/ /• 

Xf and jf =coordinates of point o', the center of pressure; 

x and y = coordinates of any point of area A BCD) 

r' = radius of gyration of area A BCD about GH) 
r" = “ “ “ “ “ “ “ EK. 

When the Center of Pressure is on an Axis of Symmetry.— 

If o' is on the axis EK, as shown in II, Fig. 25, we may replace 
F by F', an equal and parallel force acting through the center 
of gravity 0 in the same direction as F, and the couple F"F, in 
which F" is a force equal and parallel to F, acting through the 
center of gravity 0 in a contrary direction to F '. 

The stress on the fibers of A BCD produced by the force 
F' and the couple FF" must be the same as those produced by 
the force acting at o'. 

From Chapter I we know that the force F' acting at the center 
of gravity produces uniform stress over the area of cross-section, 
and hence if S\ is the unit stress at any point due to F ', 


F' F 
Sl ~A ~ A 


( 3 ! 6 ) 


The couple FF" rotates the area A BCD about some line in 
its plane. Under the hypothesis above given this axis must, 
as in bending, pass through the center of gravity of the area. 
Since the point 0' is on EK, one of the axes of symmetry, the 
axis of rotation must be GH, the other axis of symmetry. 

Since the cross-section rotates about a line in its plane, the 
unit stress at any point varies directly with its distance from that 
axis. The law of distribution must therefore be the same as 
that for bending, which is 


s'"y 


My 

I 


My 

AH 


(3U) 





I 12 


CIVIL ENGINEERING . 


in which s"'y = unit stress at any point; 

M = bending moment or moment producing rotation; 
y = distance of any point from the axis of rotation or 
the neutral axis; 

I = moment of inertia of cross-section; 

A =area of cross-section; 

r =radius of gyration of area A about the neutral 
axis. 


If in this expression we substitute s 2 for s"'y, Fxf for M , 
x for y, and r' for r, the resulting expression 


FxfX 

S2= I?z 


(318) 


gives the unit stress at any point of the area A BCD when the 
moment of rotation is Fxf and the axis of rotation is GH. 

The total unit stress 5 due to the force F' and the couple 
FF" will therefore be 


F FxfX 

S^Si+S 2 =-J + -^y2 



(3 1 9 ) 


The sign of the second term must be positive, since the couple 
FF" will produce in the area GBDH, where x is positive, the same 
kind of stress, compression, as that produced by the force F'; 
in the area ACGH, where # is negative, the stress produced by 
the couple FF" will be tension or contrary to that produced by 
F'. Positive values of 5 will therefore represent compression, 
and negative values tension. The area GBDH is the positive 
and the area ACGH is the negative part of the area ABCD 
due to the couple acting alone. 

In equation (319), if x f =o, that is the center of pressure is 

F 

at the center of gravity of the cross-section, s=— for all values 

jtL 

of x, positive and negative, or the stress over the area ABCD 
is uniform. 

This is in accordance with our hypothesis that a central load 






CONTINUOUS BEAMS, ETC. 


11 3 

in a column always produces uniform stress in every cross- 
section. 

F 

If x — o, s will be equal to — for all values of Xf. Hence 

the unit stress at the center oj gravity of a cross-section will he 

the same whether the column he centrally or eccentrically loaded. 

The stress at the center of gravity is the mean stress. 

On the positive side of the cross-section the value of 5 increases 

algebraically with ^ and Xf. If, therefore, we assume any point 

in the cross-section by making x constant, the corresponding 

_ . . d 

value of 5 will be greatest when xy is greatest or equal to —. If 

we assume the center of pressure as fixed by making Xf constant, 

d 

s will be greatest when x is equal to —. The maximum value 

of s must be when x and Xf have each their greatest values or 

d 

are each equal to —. 

2 

Hence the unit compressive stress at any point of the cross- 
section of a column eccentrically loaded is greatest when the 
center of pressure is on the same side of the center of gravity as 
the point considered , and as far from the axis of rotation as 
possible. 

For any position of an eccentric load the unit compressive 
stress is greatest at the point farthest from the axis of rotation 
on the same side of that axis as the center of pressure. 

The maximum unit compressive stress is at the center of pres¬ 
sure when the center of pressure is as far from the axis of rotation 
as possible. 

On the negative side of the cross-section, for which x is nega¬ 
tive and Xf positive, the value of s decreases algebraically with 
x and Xf. If, therefore, we assume any point in the cross-section 
by making x constant and negative, the corresponding value of 
5 will be algebraically least when x f is positive and greatest, 

or equal to +—. If we assume the center of pressure as fixed 

and positive by making x f constant, 5 will be algebraically least 

d 

where x is numerically greatest and negative, or equal to 


CIVIL ENGINEERING. 


114 

The minimum value of s must be when x and Xf have each their 
greatest numerical values but x is negative and Xf is positive, 

d d 

or x=—~ and Xf= +—. 

2 2 

The least algebraic stress may be the least positive or compres¬ 
sive stress, or it may be the greatest negative or tensile stress, de¬ 
pending on the sign of 5. 

Hence the unit compressive stress at any point oj the cross-sec¬ 
tion oj a column is least or the unit tensile stress is greatest when 
the center oj pressure is on the opposite side oj the axis oj rotation 
and as jar jrom that axis as possible. 

For any position oj the eccentric load the unit compressive 
stress is least or the tensile stress is greatest at the point farthest 
jrom the axis oj rotation on the side not containing the center oj 
pressure. 

The minimum unit compressive stress or the maximum unit 
tensile stress is at the point farthest jrom the axis oj rotation on one 
side oj that axis when the center oj pressure is as jar as possible jrom 
the axis on the opposite side. 

For any value of Xf all points having the same value of x 
have the same value of 5. Hence jor any position oj the center 
oj pressure the unit pressure is uniform along all lines parallel to 
the axis oj rotation oj the plane oj cross-section. 

To find the position of the center of pressure when the value 
of s at E is zero, or the distribution of the stress along EK is as 
shown in VI, Fig. 25, we must substitute for x in equation (319) 

—make s = o, and solve for Xf. The resulting value of x* is 

2 r f ^ Id 2 

'~t~. For a rectangular cross-section r' = \l —, hence 

12 

• • 

• ffr v 

d 

x, ~6 ‘ 

For a smaller numerical value of x f , s is positive or the stress 
at E is compressive; for a larger numerical value of x fi s is nega¬ 
tive or the stress at E is tensile. 

Hence the stress over the entire cross-section is compressive ij 
the center oj pressure is on the axis oj symmetry EK and not more 



CONTINUOUS BEAMS, ETC. 


°5 


2/ 2 

than jrom the center 0] gravity; or in the case oj a rectangular 


d 


cross-section more than — jrom the center oj gravity. 


2T 


»2 


d 


If, in equation (319), we make and x = ~> we shall have 


2F , 

S — -S' 1 —(— V2 — ~r — 2S . 

A 


• • • • ( 3 20 ) 


That is, when the unit stress is zero at one extremity oj the axis oj 
symmetry containing the center oj pressure , the unit stress at the 
other extremity is twice the mean stress. 

If the center of pressure is nearer the center of gravity, the 
distribution of the pressure is as shown in V, Fig. 25; if farther 
from the center, it is as shown in III, Fig. 25. 

d d 

If, in equation (319), we make Xf= — and x=— , we shall have 


F l d 2 
*=*i+.s2= Ji+^r 2 


(321) 


d 2 . 4 F 

In the case of a rectangle, in which r /2 = —, this becomes —r; or, 

the maximum unit stress at any point oj a rectangular cross-section 
is jour times the mean stress and oj the same character as the mean 
stress. 

If, in equation (319), we make x f=~ and x= ——, we have 


F 


d 2 


s = s 1 + s 2 - A \ y i 


In the case of a rectangle this becomes 


2 F f 

—j-; or, the minimum 


unit stress at any point oj a rectangular cross-section is twice the 
mean stress, but oj opposite character. 

When the Center of Pressure is Not on One of the Axes of 
Symmetry.—If the point o' is not on one of the axes of symmetry, 








CIVIL ENGINEERING. 


116 


the resultant F may be resolved into two components whose points 
of application are on these axes. 

These points of application must also lie on the same straight 
line through o'. 

Let F= intensity of force applied at o '; 

Xf and yf = coordinates of point o '; 

F\ and F 2 = components of F acting at o" and o'" on the axes 

of X and Y; 
x/ = abscissa of F\ ; 
y{ — ordinate of F 2 ; 

5 = unit stress at any point of surface A BCD due to F; 


c _ a << a n (( n a a it i i 77 . 

^3 r\. 


S 4 


__ t i 


it a a 


a a 


i i 


a nr? 

-T2. 


Then from equation (316) we have 

FJ x{ 

* 3 - t( i + ?2 


'x\ 

72 n 


• (322) 


hence 


F, 


s 4 = 


y/y 

r" 2 J ’ * 


5=53 + 54 


F\( x{x 


• • • • 


1 + 


r ' 2 




• (323) 


(324) 


If 0" and o'" are equally distant from o', then 


Fi=F 2 = —, x/ = 2x f , and y/ = 2y f . 
2 


Substituting we have 


t= Z/ T , , yjz 

A\ r' 2 + r" 2 F 


(325) 


d 2 £2 

For a rectangular cross-section, since r ' 2 = — and r " 2 = — this 

12 12’ 

becomes 


F( !2XfX 1 2Vfy \ 

A\ + d 2 + b 2 )' ' • 


• ■ (3 2t >) 













CONTINUOUS BEAMS , ETC. 


117 

If we determine the stress at any point on EK for any posi¬ 
tion of o', and then find the point on GH which has the same 
stress, by substituting this value for 5 in equation (326), making 
x=o, and solving for y, the line joining these points must be a 
line of uniform stress, and the line parallel to it through the 
center of gravity 0 must be the axis of rotation. The line 
through 0 perpendicular to this axis is the line of greatest varia¬ 
tion in stress. 

It follows that if the center of pressure o' is in the area 
oGBK, the greatest stress at any point of the area A BCD will 
be at B, and the least stress will be at C. 

It also follows that the stress will be compressive at C if the 
center of pressure lies between 0 and JM ; it will be zero at C 
if the center of pressure lies on JM ; and it will be tension at C 
if the center lies between JM and B. 

Distribution of Stress in Masonry Columns Eccentrically 
Loaded.—The joints of such a column offer no reliable tensile 
resistance, hence the minimum stress should never be negative. 
The stress developed by the force F must always be compressive, 
and the resultant of the total compressive stress must always 
pass through o', VII, Fig. 25, which is the center of gravity of the 
triangle of pressures, if the hypothesis is adopted that the pres¬ 
sure is always uniformly varying. 

If A BCD represents a joint in such a column, and the center 
of pressure is on the axis of symmetry EK, the distribution of 
stress along EK will be as shown in IV, Fig. 25, if the center of 
pressure is at the center of gravity; as shown in V, Fig. 25, if the 

2 Y^ ^ 

center of pressure is between the center of gravity and and 

2 r ' 2 

as shown in VI, Fig. 25, if at a distance from the center of 
gravity. 

2 r'* 

If beyond —from the center of gravity, the distribution 

will be as shown in VII, Fig. 25, the point of zero pres¬ 
sure on EK being twice the distance from 0' that K is 
from o'. 

Since the total compressive stress remains constant as 0' 
moves towards K, the pressure midway between K and the 





ii 8 


CIVIL ENGINEERING . 


F 


point of zero pressure will be —, and the pressure at K will be 
2 F . . 

—, in which ^4' = area between and the limiting line of zero 


pressure. 

As o' approaches K , the area A' will decrease; and finally 

F 

when o' reaches its limit K , -77 will be infinite. 

’ A' 

As it is desirable to have the pressure over a masonry joint 
as uniform as possible, the condition is usually imposed that 

2/2 

the center of pressure shall never be more than —j~ from the center 

of gravity of the joint, or in a rectangular joint the center of 
pressure must never be more than one-sixth the depth of the 
joint from the center of gravity. 

In masonry joints the resultant pressure is often oblique to 
the joint. The center of pressure of any horizontal joint is, 
however, the point where this resultant pierces it. If the result¬ 
ant pressure is oblique, F becomes the component of the result¬ 
ant pressure normal to the surface of the joint. 


PROBLEMS. 

40. The base of a block is 7X10 feet, and the center of pressure 
is one foot from the center of the base, measured parallel to the 
longer side. The total pressure being 40,000 pounds, find the 
unit pressure at the corners. 

A ns. 228.6 and 914.3 pounds per square foot. 

41. At the base of a wall 6 feet wide the pressure at one edge 

is 20 pounds per square inch; at the opposite edge it is 5 pounds 
per square inch. What is the mean pressure and where is the 
center of pressure? Ans. 12.5 pounds per square inch. 

Xf =7.2 inches. 



CHAPTER VI. 


MOVING OR LIVE LOADS ON A BEAM. 

Bending Moments.—A load which moves over a structure and 
is not continuously supported by it in the same position is called 
a live or moving load. It is concentrated if it acts at a single point; 
it is uniformly distributed if it is either a continuous load of uni¬ 
form weight or made up of many equal concentrated loads sepa¬ 
rated by short intervals. A railway train is a live or moving load 
on a railway bridge. The wheel loads are concentrated loads; 
the entire train is often treated as a uniformly distributed load. 
The live load on each elementary part of the structure is so much 
of the entire live load as is transmitted through it. 

A load which is continuously supported by a structure is its 
dead or permanent load. The weight of the superstructure and 
track is the dead load of a railway bridge. The dead load is 
usually treated as a uniformly distributed load, but may also be 
treated as a number of concentrated loads. The dead load in 
each elementary part of a structure is so much of the entire dead 
load as is transmitted through it. 

Concentrated Live Loads.—I. If a single concentrated load 
is moved over a beam without weight resting on end supports, the 
bending moment at any section of the beam will be numerically 
greatest when the load is at the section. Thus in the beam shown in 
Fig. 26 the bending moment at the cross-section C will be great¬ 
est when the weight is at C. 

Let clockwise moments be positive. 

0 = the origin of coordinates; 

1 =length of the beam; 

# = distance of center of gravity of load from O; 

d = “ “ section from O ; 

W = weight of concentrated load; 


' 120 


CIVIL ENGINEERING. 


= reaction at 0 = 


W(l-x) 

l 


M' = bending moment at C ; 

If" = “ “ ‘ ‘ the load; 

M m — maximum bending moment in beam due to live load. 


i 



Fig. 26. 


1. While the load moves jrom O to C we have 

. iv 

Wd(l-x) 

M' = Rid — W(d—x) = -—- -W(d-x) = Wx 



( 3 2 7 ) 


Since d<l, M' must be positive and will increase with x, until # 
reaches its maximum value, 4 , when we shall have 


M' = W d - j— .(328) 

2. While the load moves jrom C to X we have 

M'=R!d=WdXXI= W d(i-^. . . . (329) 

Since x<l,M' will be positive and will decrease while # increases, 
until x reaches its maximum value, /, when we shall have 

M ' = o .(330) 

Therefore the live-load bending moment at the section C is great¬ 
est when the live load is at C. 













MOVING OR LIVE LOADS ON A BEAM. 


12 i 


II. The dangerous section is the section at the middle point of 
the beam. 

If, in equation (328), we substitute x for d, M' becomes M", 
the bending moment at the load, and 

Wx 2 

M -Wx-—j— .(331) 

If we differentiate this equation with respect to x, and place 
the first differential coefficient equal to zero, the resulting value 
of x will be the distance of the concentrated load from O, when 
the bending moment at the load is a numerical maximum. 

dM" 2 Wx l 

-^r= w —r=°’ whence *-j- • • (332) 

The bending moment at the load is therefore a maximum when 
the load is at the middle point of the beam, and 


M = 

m 


Wl 

4 


( 333 ) 


These same conclusions are shown graphically in Fig. 26, 
where the parabola OWX is the curve constructed from equation 
(331), and the broken line OWX is the line of bending moments 
from equations (327) and (329), if d is variable and x constant. 

Since the broken line lies below the parabola at every point, 
except W, for every position of the load, it follows that the bending 
moment at C is greatest when the load is at C, and its bending 
moment is represented by the ordinate of the parabola. 

Since the ordinate of the parabola at any point represents 
the greatest bending moment at that point, it follows that the 
middle point, where this ordinate is greatest, will be the dangerous 
section if the beam is of uniform cross-section. 

III. Propositions I and II are equally true ij the beam is of 
uniform cross-section and weight , and its weight is considered; or 
if we combine the live with a uniform dead load. 

The resulting bending moment at any cross-section is the 
sum of the bending moments due to the dead and live loads 







122 


CIVIL ENGINEERING . 


acting separately. Since the bending moment at any section 
due to the dead load is independent of the position of the live 
load, their sum will have its greatest value for any section when 
the bending moment at that section due to the live load alone 
has its greatest value, or when the live load is at the section 
considered. 

Since the bending moment at the middle point of the beam 
is a maximum for both live and dead loads separately, it must be 
a maximum for their sum. This will therefore be the dangerous 
section. 

IV. If two unequal concentrated loads, separated by a fixed 
distance, move over a beam without weight resting on end sup- 


Y 


% 1/' r* ' •/ y/ 



Fig. 27. 


ports, the dangerous section will be at the greater load when that 
load is as jar jrom one support as the resultant oj the loads is jrom 
the other. 

In Fig. 27 let F"=greater load, 

W" = lesser load, 

F = resultant of W' and IF", 

O = origin of coordinates, 
d" = distance of F from O, 


d' = 

1 < 

“ F 

(( 


a = 

< i 

“ F 

11 

W', 

b = 

11 

“ F 

1 c 

W", 

od = 

< < 

“ W' 

< < 

0, 

x" = 

< ( 

“ \y " 

C ( 

X, 


Fd' 

Ri = reaction at 0 = —y-, 


Ro= 


(( 


< c 


X 


F(l-d') 

l 


Fd" 

l 1 











MOVING OR LIVE LOADS ON A BEAM. 


123 


M' = bending moment at W', 

M"= “ “ “ W". 

To prove the above proposition, it is necessary to show— 

1 st. That the bending moment at one of the loads is equal 
to or greater than the bending moment at any other section of 
the beam. 

2d. That the bending moment under each load is a maximum 
when the load is as far from one support as the resultant of the 
two is from the other, or when the middle point of the beam 
bisects the distance between the load and the resultant. 

3d. That the maximum value of M ', or M J, is greater than 
the maximum value of M", or M m ". 

1. If we construct the line of bending moments for any posi¬ 
tion of the two loads, as that shown in Fig. 27, it will be the 
broken line OABX. For different positions of the load, the 
line AB may be horizontal or slope towards O or X. That is, 
the bending moment at one or the other of the loads will be equal 
to or greater than the bending moment at any other section of 
the beam. This might also be shown analytically. 

2. To prove the second statement, for the section at W' we 
have 

Fx'd' 

.( 334 ) 

But d! = l—od — a, hence 

M' = -x' 2 -ax f ) .(335) 

If this expression is differentiated with respect to X and its 
first differential coefficient is placed equal to zero, the value 
of X deduced will be the distance of W' from O when M' is a 
maximum. 

dM' F 

__ =7(/ _ 2 ^_ a) .(336) 

Making the first differential coefficient equal to zero we have 

F l —a 

—(I-2X-a)=o, or X = - 


2 


• • ( 337 ) 








124 


CII/IL ENGINEERING 


Therefore when W' is at a distance of-from O, M' is a maxi- 

2 

mum. Substituting this value of x' in the value of d' given be¬ 
low equation (334), we have 

d' = l-x’-a = {l-a) - • • • (338) 

which is the same as the value of x / . Hence when M' is a maxi¬ 
mum, x' = d r , or W' is as far from 0 as F is from X. 

If the origin is taken at X, we have for the bending moment 
at W" 


M" = R 2 x" 


Fx"d" 

l 


But d" = l—x" — b, hence 


(339) 


F 


M" =-j(lx" —xf ’ 2 — bx"), 


( 340 ) 


SM" 

dx" 


-r(l — 2x" — b). 


o ( 341 ) 


Making 


dM" 

■ ^ ,r = 0 we have 
ox 



Therefore when W" is at a distance- from X, M" is a maxi- 

mum. 

Substituting this value of x" in the value of d" above, we have 
l-b 

d n =-. Hence when M" is a maximum, x" = d", or W" is 

as far from X as F is from O. 

3. To compare the maximum values of M' and M”, we must 
substitute for x f and x", in equations (334) and (339), the values 










MOVING OR LIVE LOADS ON A BEAM. 


125 


which correspond to the maximum values of M' and M". We 
shall then have 


Fd ' 2 Fd " 2 

— j~ an( ^ ^ . . . . ( 34 2 ) 


Since a<b, the value of d ', when the middle point of the beam 
bisects a , will be greater than the value of d" when the middle 
point bisects 6; hence the maximum value of M' is greater than 
the maximum value of M". 

Therefore when the greater load is as far from one support as 
the resultant is from the other, the section at the greater load is 
the dangerous section. 

Uniformly Distributed Live Load.—V. If a live load of uni¬ 
form weight moves over a beam without weight, resting on end 
supports, separated by a distance less than the length of the live 
.oad, the bending moment at every section oj the beam will be great¬ 
est when the live load entirely covers the beam. 



In Fig. 28 let 
l = length of beam; 
a = or >1 = length of moving load; 
d = distance of section C from O; 

_ 11 << it (( (( • 

x = loaded part of beam when load moves on; 



(( a a n a 


( c 


off. 


Ri = reaction at 


O due to moving load = 




< ( 
















126 


CIVIL ENGINEERING. 


w = weight of moving load per unit of length; 

M' = bending moment at C when D is between O and C ; 
M J = greatest numerical value of M ’; 

M" = bending moment at C when D is between C and X ; 
M m " = greatest numerical value of M"; 

M m = maximum bending moment in beam due to live load. 


i. While the head oj the live load moves jrom O to C. 
bending moment at C is 

/ N wxdll—— 

M' = R\d—wx(d—^\ =- -wxld-* 


The 


wx 2 d . wx 2 vox 2 1 

= wxd — r- — wxd +-=- I 

2 1 2 2 \ 



( 343 ) 


Since d<l, M f is positive and increases as x 2 increases, and will 
be zero when x is zero, and greatest when x = d. Substituting 
d for x in equation (343) we have for the bending moment at 
C when the head of the load is at C, 



wd d , 
2 l v 


(344) 


2. While the head oj the moving load moves jrom C to X . The 
bending moment at C is 


M" = Rid 


wd2 wx 
2 l 


wd 2 

2 


wxd 


wx 2 d 
2 1 


wd 2 

—■ (345) 


If this equation is differentiated with respect to x , and the 
first differential coefficient is placed equal to zero, the resulting 
value of x will give the distance of D from O when j|£" is a 
maximum: 


dM " 
dx 


= wd 


wdx 


l 


= 0. 


• • (346) 














MOVING OR LIVE LOADS ON A BEAM. 


127 


Solving for .v we have x = l. Substituting this value in equation 

(345) we have 

M m "^(l-d) .( 347 ) 

M m " must be greater than M J, since MJ is equal to M m " 

d 

multiplied by the factor y, which is less than unity. 

To find the value of M m , the maximum bending moment for 
any position of the load, we must make d in equation (347) vari¬ 
able and find the maximum value of MJ f : 


dM " wl 
_^- =— wd=o. 
Od 2 


• (348) 


Hence d=— corresponds to a maximum value of M m ". 
2 

Substituting this value in equation (347) we have 


wl 2 


( 349 ) 


The middle section is therefore the dangerous section. 

If the origin is taken at X, and the distance from X to the tail 
of the load is represented by od, similar expressions to those given 
above may be obtained for the bending moment at C, as the tail 
of the load moves from O to C, and from C to X. 

From these expressions it may also be shown as above that 
the bending moment at C is a maximum when the live load 
covers the entire beam. 

These results are shown graphically in Fig. 28. 

The parabola OBX is the curve of bending moments when the 
load covers the entire beam. 

The parabola OD and the straight line DX together form the 
curve of bending moments for the live load extending from O to D. 

It is evident that if x = o, the line ODX will coincide with OX, 
and if x = l, the line ODX will coincide with the parabola OBX. 

Hence the maximum bending moment M m " at C must be the 
ordinate of OBX at C; and the maximum bending moment of 







128 


CIVIL ENGINEERING. 


the beam or M m will be the ordinate of the parabola OBX at the 
middle point. 

VI. Proposition V is equally true ij the beam upon which the 
live load is moved is oj uniform cross-section and weight , and its 
weight is considered , or if we combine the live with a uniform 
dead load. 

The resulting bending moment at any cross-section is the sum 
of the bending moments due to the dead and live loads acting 
separately. Since the bending moment at any section due to the 
dead load is independent of the position of the live load, their 
sum will have its greatest value for that section when the bending 
moment due to the live load alone has its greatest value or when 
the live load covers the entire beam. 

Since the bending moment at the middle point of the beam is 
a maximum for both the live and the dead load separately, it 
must be a maximum for their sum. This will therefore be the 
dangerous section. 

VII. If the uniform live load is shorter than the beam upon 
which it moves, the maximum bending moment at any section will 
be greatest when the section divides the load into segments whose 
lengths are proportional to the lengths of the segments of the beam 
on either side of the section. If the segments of the load and 
beam to the left of C are L! and and on the right of the section 
L" and l", then will the bending moment at C be a maximum 
when L r :l'::L": l". This proposition may be proved in a man¬ 
ner similar to the other proofs. 

PROBLEMS. 

42. A wheel weighing 500 pounds is rolled over a beam 20 
feet long whose weight is 50 pounds per linear foot. Find the 
maximum bending moment in the beam in inch-pounds. 

Ans. 60,000 (inch-pounds). 

43. Two carriage-wheels, separated by a fixed distance of 
8 feet, move over a beam without weight, 25 feet long. The load 
on one wheel is 3000 and on the other 2000 pounds; find the maxi¬ 
mum bending moment in the beam. 

Ans. 23,762 (foot-pounds). 

44. A beam 25 feet long, weighing 50 pounds per linear foot, 
is subjected to a moving load 30 feet long whose weight per linear 
foot is 100 pounds. Find the greatest bending moment at a section 
of the beam 10 feet from either support. 

Ans. 11,250 (foot-pounds). 


MOVING OR LIVE LOADS ON A BEAM. 


129 


Shearing Stress. 

Concentrated Load.—VIII. If a concentrated load is moved 
over a beam without weight resting on end supports, the shear at 
any section will be numerically greatest when the load is at the 
section . 



In Fig. 29 let O = origin of coordinates; 

/ = length of beam; 

w = weight per unit of length of beam; 

W = weight of concentrated load; 
d = distance of any cross-section from O; 
x — distance of concentrated load from O; 

W(l-x) 

Ri = reaction at 0 = — -j— for live load alone, 
W(l — x) wl 

= -—— + — for both live and dead loads; 

V" — shear due to the concentrated load, or the 
live-load shear; 

V s = resultant shear of live and dead loads. 

Considering the live load alone, we have for any section be¬ 
tween O and J, as the one marked C', 


Vs" = 


IV (l-x) 

l 5 


( 35 °) 


which is positive since x<l. In this expression V/' will be great 1 



















1 3 ° 


CIVIL ENGINEERING. 


est when x is least, that is when x=d; or the weight is at the 
section considered or is moved back to C'. 

For any section between J and X, as the one marked C, we 
have 


,_ W{l-x) Tr Wx 


l 


l 


( 35 1 ) 


which is negative. 

This value of V 8 " increases numerically as x increases, and is 
greatest when x = d or the load moves to C. Hence the shear at 
any section due to the live load alone is a maximum when the 
concentrated weight is at that section. 

This is shown graphically in the figure. If the line of live- 
load shear DEFG is plotted from O to X, its ordinates will be 
positive between O and J, and negative between J and X. The 
positive ordinate OD increases as the weight W is moved towards 
O, and the negative ordinate XG increases as the load is moved 
towards X. 

The maximum shear at the supports is greater numerically 
than the maximum shear at any other section, since the expression 

W(l-x) IV x 

F/' =- j -is greatest when x=o, and V 8 " = - j- is greatest 


when x = l. 

IX. If a concentrated load moves over a uniformly loaded 
beam, the dead- and live-load shears will have like signs between 
the load and the nearer support , as well as between the middle point 
oj the beam and the farther support; but unlike signs between the 
load and the middle point. 

For any section between O and W, as the one marked C', Fig. 
29, we have 



( 35 2 ) 


In this expression wy-^ — d'j is the shear due to the dead load, 

and w(^i— yj is the shear due to the live load. If we consider 

lx 

only the part of the beam between O and J, —>d and 1 >—, their 

2 l 






MOVING OR LIVE LOADS ON A BEAM. 


signs are both positive and their sum is numerically greater than 
either. This is shown graphically in the figure. At any section 
between O and J the ordinates of HI , the resultant line of shear, 
are greater than the ordinates of DE or AB, the component lines 
of shear. 

For any section between B and X, as the one marked C, we 
have 

/ / \ Wx 

V s = R 1 -wd-W = 'wl — -d)-—, . . . (353) 


(l \ . Wx 

in which wl— — dj is the shear due to the dead load, and —— 

1 

is the shear due to the live load. Since —<d, both terms of the 

2 

second member are negative, hence the resultant shear at the sec¬ 
tion is negative and is numerically equal to the sum of the shears 
of the dead and live loads. 

This is shown graphically in the figure. At any section be¬ 
tween B and X the ordinate of the resultant line of shear, JK , 
is equal to the sum of the ordinates of BL and FG, the component 
l'nes of shear. 

For any section between J and B, as the one marked C", we 
have 

/ / \ Wx 

V s = Ri-wd-W=w^—-dJ—j-. . . . (354) 

Since —>d, the first term of the second member, which is 
2 

the shear due to the dead load, is positive. The resultant 

(l \ x 

shear will be positive if w[——dj>W-j\ it will be zero when 

wl - dj=W-j, or when at any section the shear due to the 

live and concentrated loads is equal; it will be less than zero 

(l \ Wx 

or negative when — d)<—, or the shear due to the 

dead load is numerically less than the shear due to the live 
load. 

This is shown graphically in Fig. 29, in which ABL is the line 





i3 2 


CIVIL ENGINEERING. 


of shear due to the dead load, DEFG is the line of shear due to 
the live load in the position shown, and HIJK is their resultant 
line of shear. The ordinates of HIJK are numerically greater 
than those of ABL between O and J and between B and X ; and 
algebraically less than those of ABL between B and J. The 
resultant line passes through zero, the point of zero shear being 
either at W or between W and B. 

If we draw the line OM, whose ordinate XM = EF, it is the 
locus of the point F as the live load moves from O to X ; its ordinate 
at any point, as C", will give the negative live-load shear between 
C" and X when the load is at C". Since there can be no section of 
zero shear to the left of the load W when the load is to the left of 
B, the point where the ordinate of AB is equal to that of OM is the 
farthest limit to which the point of zero shear can move from the 
middle point to the left. This distance depends upon the relative 
values of w and W as shown by the value of d deduced from the 

; by making x = d, 
d 

2 IT T 2Wl 


. (355) 

\ 


(l \ Wx 
equation w\— — d\ = — 


There will be a corresponding point on the opposite side of 
B equally distant from it, since, when the load is at C, the live- 
load shear between B and C will be positive and the dead-load 
shear negative. These two points will fix the limits between 
which the shear at every section must change its sign as the load 
moves from O to X. 

The resultant shear V s at any section, due to the combined 
moving and dead loads, will be a numerical maximum when the 
load is at that section. This arises from the fact that the dead¬ 
load shear at any section is constant, and the live-load shear is a 
maximum of the same sign as the dead-load shear when the load 
is at the section. 

When the live load is at the section adjacent to either support, 
the shear at that section will be the maximum shear in the beam, 
since the ordinate OH of the resultant shear will then have its 
maximum value. 

Uniformly Distributed Live Load.—X. If a uniform live load 
is moved over a uniformly loaded beam, resting on end supports 





MOVING OR LIVE LOADS ON A BEAM . 


133 


separated by a distance less than the length of the live load, the 
shear at any cross-section has one limiting value algebraically when 
the live load covers the greater segment into which the section divides 
the beam, and another when the load covers the smaller segment. 



In Fig. 30 let / = length of beam; 

a = total length of moving load, =or>/; 
w = weight per unit of length of beam; 
w f = weight per unit of length of live load; 

l 

d = distance to any section, as C, at which d<—, 
from O; 

x = distance of head of live load from O; 
x' = distance of tail of live load from O; 

Ri — reaction at O ; 

R 2 = reaction at X; 

V s r = dead-load shear at C ; 

V " = live-load shear at C ; 

V s = V s ' + V s " = resultant shear at C. 

Then will V s be a minimum when x = d, and a maximum when 
x' — d. 

1. While the head of the live load moves from 0 to C we have 
















134 


CINIL ENGINEERING. 


V s =Ri~ wd — w'x 


'h x \ 

, w'x\l——) 
wl \ 2 ) 

T + ~l 


— wd — w'x 


= w 




w'x 2 
2 / 


( 35 6 ) 


The first term of- the second member is the dead-load shear 
at C which is constant for all values of x } and is positive since 

1 

— >d. The second term is the live-load shear at C which in- 

2 

creases numerically as x 2 increases. The resultant shear will there¬ 
fore decrease algebraically as *v increases, and will have its greatest 
value when x = o, and its least value when x = d. 

When x = o, 

F «=w(j-<^ = W;.(357) 

when x = d, 

11 \ w'd 2 

V s =w(j-d)- — = V.’-V.". . . . (358) 


In equation (358) T s may be positive, negative, or zero, depend 

l -d 


ing on the relative values of w 


and —t ; the relative 


2/ 


w 

values of these quantities depend upon the ratio —. The mini¬ 
mum value is not a numerical minimum, unless the value of V s 
in (358) is positive or zero. 

2. While the head of the load moves from C to X, 


V S = R\ — (w + w')d 


wl 

2 



(w+w')d 


=w[ — — d)+w' 



■ (359) 









MOVING OR LIVE LOADS ON A BEAM. 


1 S5 


The first term of the second member is the dead-load shear 
at C and is identical with the first term of the equation (356). The 
second term is the live-load shear at C and increases algebraically 
as x increases, since its first differential coefficient is positive. 
The resultant shear will therefore increase algebraically as x in¬ 
creases, and will have its least value when x = d, and its greatest 
value when x = l. 

When x = d, 


V 8 =w 




w'd 2 

IT’ 


(360) 


When x = /, 


V 8 =w[— — aj = (w-\-w')y^ — d ). . (361) 


The value of V s in equation (360) may be positive, zero, or 
negative, depending upon the relative values of w(^— — d^j and 
d 2 

w '—; the value of V s in equation (361) will always be positive, 
2 

since —>d. 

2 

3. While the tail of the load moves from O to C, 


V s = Ri —wd—w'(d—x') 


wl w'(l-x ') 2 
2 + 2 1 


— wd — w' (d — x') 





( 3 6 2 ) 


The first term of the second member is the dead-load shear at 
C and is identical with its values in the preceding equations. The 

l 

second term is the live-load shear; and since — >d, it is positive 

and increases with x'. The resultant shear therefore increases 
with xf, and is least when x'=o, and greatest when x'=d. 






136 


CIVIL ENGINEERING. 


When x' =o } 

V s = (w+w')(^-d S J\ .(363) 

when x' = d, 

V„=w(j- d) +w'[j-d+. . . . (364) 

Since — >d, the values of V s given by equations (363) and (364) 
are both positive. 

4. While the tail of the load moves from C to X, 

V s = Ri —wd 

wl w r (l — x / ) 2 

=— +-7— --wd 

2 2/ 




• (365) 


The first term of the second member is the dead-load shear 
and is identical with its values previously given. The second 
term of the second member is the live-load shear and decreases as 
od increases, since its first differential coefficient is negative. The 
resultant shear will therefore decrease as x' increases, and will be 
greatest when x f = d, and least when x f = l. 

When x =d, 

v *=™(l-d)+’ w '(l-d+ Jj); . . . (367) 

when x' = /, 



Since— >d, the values of V s in equations (367) and (368) are both 
positive. 

Hence we see that the resultant shear at C decreases alge¬ 
braically while the head of the live load moves from O to C; it 
increases algebraically while the head of the load moves from C to 
X and while the tail of the live load moves from O to C, and 





MOVING OR LIVE LOADS ON A BEAM. 


137 


finally decreases while the tail of the live load moves from C to X. 

1 

Had the section been taken beyond B , as at Q , so that — <d, the 

2 

maximum resultant shear would have been when the head of the 
load was at Q , and the minimum when the tail was at Q. 

Section of Zero Shear.—If, in equation (356), we make F s = o, 
d will be the distance from O to the section of zero shear. Solv¬ 
ing with respect to d we have 

/ w' x 2 
2 w 2I ’ 


If we assume x = o } the value of d will give the position of the sec¬ 
tion of zero shear before the live load comes on the beam. If 

x=o, d=—, or the section of zero shear is at the middle point of 
2 

the beam. 

As x increases, d decreases, and the section of zero shear ap¬ 
proaches O. When x = d, its maximum value in equation (356), 
the section of zero shear is at the head of the live load and 


Hence 


or 



w'd 2 
w 2 1 


w 

d = l —d — i i 
w y 



d 

l 



The ratio of -j depends therefore on the ratio — r 


(369) 


(370) 


If, in equation (359), we make V s = o, d will also be the dis¬ 
tance from O to the section of zero shear. Solving with respect 
to d we have 


wl 


w'x 2 

w'x — - T 

21 


d = 


2 


w + w' 


(371) 








! 3 8 


CIWL ENGINEERING. 


This value of d increases as # increases, hence it will be least when 
x = d and greatest when x = l. Making x = d, we have as before 



(372) 


Making x = l, we have d = —. 

2 

These last equations show that as soon as the section of zero 
shear meets the head of the live load, the section of zero shear 
moves towards the middle point of the beam and reaches the 
middle point when the live load covers the entire beam. 

As the head of the live load therefore moves from O to A^, the 
section of zero shear moves from B and meets it at some point 
between B and O, whose distance from O depends upon the ratio 


w 


—; it then moves back to the middle point of the beam. 
w 

In a similar manner it may be shown that while the tail of 
the load moves from O to X, the section of zero shear moves to a 
corresponding point between B and X , which it reaches simul¬ 
taneously with the tail of the live load; it then moves back to the 
middle point. If we substitute for w and w' their values in equa¬ 
tions (370) and (372), we may find the distance from O and X to 
the extreme limits of the position of the section of zero shear; 
between these limits the shear at every section must change its 
sign as the live load moves over the beam. 

The effect of a live load of uniform weight may also be shown 
graphically. 

In Fig. 30 let 

0 = origin of coordinates, and OX and OY = coordinate axes; 
wl 

OA = —; 

2 

_ r 

ON = —; 

2 

A BL = dead-load line of shear from equation V 8 ' =w(^~ —ocj; 
NBM = live-load line of shear when live load extends from O to 
A" from equation V s " =w'(^— — x^J; 



MOVING OR LIVE LOADS ON A BEAM. 


*39 


OEM = parabola whose ordinates give live-load shear at the head 

— r w^ x 2 

of the live load, from equation V s " =-—; (see 

2 i 

equation (356);) 

NQX = parabola whose ordinates give the live-load shear at tail 

/1 x' 2 \ 

of live load from equation V s " =w' (— — x' 


(see equation (362);) 


2/ 


DEF = live-load line of shear when head of live load is as shown 
to the left of B\ OD = reaction at O, XF= reac¬ 
tion at X ; 

PQ T = live-load line of shear when the tail of live load is as 
shown to the right of B\ OP = reaction at O, XT = 
reaction at X. 



The locus of the point E is the parabola OEM , and the locus 
of the point Q is the parabola NQX . 

When the head of the live load is at O, the point E is also at 
O, and the line EE coincides with OX. The live-load shear is 
therefore zero at every point, and the resultant shear at any point 
is given by the ordinates of the line ABL; at the point C the re¬ 
sultant shear is positive. 




















140 


CIVIL ENGINEERING. 


When the head of the live load is at J, as shown in the figure y 
the ordinates of the lines ABL and DEF at some point between 
J and B are equal and the resultant shear at this point is zero. 
The resultant shear at any other point is found by taking the 
algebraic sum of the ordinates of lines ABL and DEF as in 
GJI ; at the point B the resultant shear is negative. 

When the head of the live load is at X or the tail at O, the 
resultant shear at any point is found by taking the algebraic sum 
of the ordinates of ABL and NBM ; at the point C the resultant 
shear is positive, and at B zero. 

When the tail of the live load is at Q, as shown in the figure, 
the resultant shear at any point is the algebraic sum of the ordi¬ 
nates of the lines ABL and PQT. 

When the tail of the live load is at X , the live-load shear at 
every point is zero and the resultant shear at any point is the 
ordinate of the line ABL. 


PROBLEMS. 

45. A beam weighing 50 pounds per lineal foot rests on end 

supports 25 feet apart; a wheel weighing 500 pounds is rolled 
across it. Find the maximum and minimum shear at the section 
10 feet from left support. Ans. Maximum +425 pounds. 

Minimum — 75 “ 

46. A beam weighing 50 pounds per lineal foot rests on end 
supports 25 feet apart and also supports a uniformly distributed 
load which is 30 feet long and weighs 40 pounds per lineal foot 
which moves over the beam. Find the maximum and minimum 
shear at the section 10 feet from left support. 

Ans. Maximum +305 pounds. 

Minimum + 45 “ 


CHAPTER VII. 


COLUMNS. 


A column is a strut whose length is such that it will perceptibly 
bend or buckle under a compressive jorce bejore it actually breaks. 
This takes place when the ratio of length to the least dimension 
oj cross-section has a certain value depending upon the character 
of the material. This ratio varies from 5 to 20 for different 
materials. 

It is assumed that in a column, when the unit compression 
reaches a certain limit, the axial fiber will be deflected in a man¬ 
ner similar to the mean fiber of a beam supporting a load con¬ 
centrated at its middle point. 

If the ends of a column are square, that is the end faces are 
perpendicular to the axis, the curve of the axial fiber will be sim¬ 
ilar to the curve of mean fiber of a beam fixed at both ends. It 
will be tangent to the original axis at the ends 
and will have two points of inflection, one 
half-way between each end and the middle 
point (A, Fig. 31). If one end of the column 
is square, and the other round or held by a 
pin, the curve of the axial fiber will be similar 
to the curve of the mean fiber of a beam, hav¬ 
ing one end fixed and the other resting on a 
support. It will be tangent to the original axis 
at the square end and there will be but one 
point of inflection, which will be about one- 
third of the height of the column from the 



Fig. 31. 


square end ( B , Fig. 31). If the column has two round or pin- 
connected ends, the curve of the axial fiber is similar to the curve 
of mean fiber of a beam resting on two supports. It has no points 
of inflection ( C, Fig. 31). 

Euler’s Formula.—The first attempt to deduce a rational 
formula for the breaking load of a column of uniform cross- 














142 


CIVIL ENGINEERING. 


section was made by Euler. He assumed a bent column with 
round ends and ascertained the least intensity of the force which, 
acting in the line of the original axis, would keep the axis of the 
column in a bent condition. He assumed this as its breaking 
weight. 

In C, Fig. 31, let 

W ,n = weight in pounds which placed on the column C will 
hold it in its bent condition; 

/ = length of the column in inches; 
y w = maximum deflection of the axial curve in inches; 
x and y = coordinates of any point on the axial curve. 

The axis of X coincides with the axis of the column before 
deflection, and the origin is at one end. 

Then we have for equilibrium between the moments of the 
external force and the internal stresses 

EId 2 y 

^2 — — W y , .(373) 


in which the first member is the moment of resistance and the 
second is the moment of flexure. Multiplying by 2 dy we have 


Integrating, 


El (2dyd 2 y) 
§x 2 


2W'"ydy. 


El dy 


dx 2 


— W'"y 2 + C. 


Where y = y m , 



hence C = W"yJ. 

y m 


Substituting, 


(374) 

(375) 


ElSf 

dx 2 


= W'"y m 2 -W , ”y 2 . 


(376) 


Solving with respect to dx 2 we have 


dx 2 =~ dy ~ 


W"'y m 2 -y 2 ' 


Extracting the square root, 


dx = ^ 


El 


ov 


W'" vV^-v 2 * 

y m y 


(377) 


(378) 













COLUMNS. 


143 


Integrating, 


El . y 

■ v —\ tt/ 777 sin 1 ~+C.( 379 ) 

y m 

Where x = o, y = o, hence C' = o. Transposing and taking the 
sine of both members, we have 


sine 



(3 8 °) 


which is the equation of a sinusoid. For x = l, y = o, hence 


W'" 

sine ( l vl-gy I = 0. • 


• • 


• (381) 


The arc whose sine is zero is n or some multiple of 7r. Hence 


W 


nr 


= or some multiple of n. ... (382) 


The least value of W'" is derived from solving equation (382), or 


W"' = 


Eln 2 AEr 2 n 2 
~1T~ i 2 


(383) 


This is Euler’s formula for the breaking load of a column with 
round ends; in it H=area of cross-section, and r = least radius 
of gyration of cross-section about an axis through its center of 
gravity. 

It will be observed that the value of W" is independent of y m9 
hence a bent column, with any deflection, must be considered 
unsafe. 

To determine expressions for the breaking weight of columns 
with square ends or with square and round ends, it was assumed 
that their breaking loads were inversely proportional to their max¬ 
imum deflections, or to the maximum deflections of their corre¬ 
sponding beams. The maximum deflections of the correspond¬ 
ing beams are given in the table, page 86, under beams loaded at 
the middle point. The maximum deflection of the beam fixed at 

1 WP 

-frr> that of a beam fixed at one end only is 

192 El 


both ends is 












144 


CIVIL ENGINEERING. 


I IT/ 3 

—- —-7and that of a beam simply resting on its supports is 
108 hi 

1 IT/ 3 

— —r. These deflections are to each other as £:£:i. 

48 EL 

Recent experiments have caused J, and 1 to be substituted 
for the former values, J, f, and 1. 

If W' = breaking load of column A , 

IT" = “ “ “ “ B, 

IT'" = “ ‘ ‘ ‘ ‘ “ C, then 


2 Elir* 2AErH r 2 

IT': IT'":: 1: J, or IT' = 2IT'" = ——, 


4 77 /■—2 4 4 77^2^2 

W":W"':: i:f, 0 r IF" = -f IP" = ^77" 3 * 


/ 2 


As Euler’s formulas are based on the hypothesis that the col¬ 
umn yields by bending alone, they are more applicable to long 
columns than to short ones. 

Gordon’s Formula.—This empirical formula credited to both 
Tredgold and Gordon is of the following form: 


w'=- s ' A 


1 + 


q'l 2 ’ 

b 2 


• (386) 


in which IT' = breaking load of A, the square-end column, in 

pounds; 

A =area of the cross-section in square inches; 

/ = length of the column in inches; 
b = least dimension of the cross-section in inches; 
s c ' = modulus of crushing of the material; 
q' = a coefficient, determined by experiments on square- 
end columns. 

The values deduced by Gordon for s/ and q' were from ex¬ 
periments made by Hodgkinson about 1840, and are no longer 
employed. 

Rankine’s Formula.—Gordon’s formula for columns with 
square ends was modified by Rankine, who substituted for b its 











COLUMNS. 


145 


value in terms of r, the least radius of gyration of the cross-section. 
The formula then became 


W'-* :A 


I + 


ql 2 * 


(3S7) 


For a column of rectangular cross-section r 2 = — 

& A 


= T V& 2 , hence 


for a solid column of this cross-section q=\^q'. 

Rankine’s formula may be deduced from the formula giving 
the greatest unit stress in a beam due to a force of compression 
and a force of flexure in the following manner. In A, Fig. 31, let 
W' = weight in pounds which will break the column; 
s' = modulus of crushing of the material; 

A =area of cross-section in square inches; 

M = bending moment in inch-pounds at the middle point; 

1 = moment of inertia of the cross-section about an axis 
through its center of gravity; 

# and y = coordinates of the axial fiber of the column; 

y' = distance of extreme fiber in the cross-section from the 
neutral axis. 

From equation (298) we have 



+ 


My' 

I 


( 388 ) 


Substituting for M its value at the dangerous section, W'y m , in 
which y = maximum deflection of the column, we have 

y m 


W' W'y y' 

- / __I 

s ‘ A + I 


(389) 


Substituting as before for I its value Ar 2 , in which r is the least 
radius of gyration of the cross-section about the neutral axis, we 
have 


W' W'y m y' 

~r 


A 


Ar 2 


or W' = 


s c 'A 


1 + 


r 2 


r 


( 39 °) 


In this formula we have the value of W' in terms of the max¬ 
imum deflection, and quantities depending on the character of 














146 


CIVIL ENGINEERING. 


the material and upon the form and dimensions of the cross- 
section. If we can obtain the value of the maximum deflection 
in terms of the length, and substitute it in this formula for y m , 
we shall have a working formula applicable to columns of all 
materials, of all lengths, and of all dimensions and forms of 
cross-section. To obtain such a value of y m we assume that the 
ratio of the maximum stress to the maximum deflection is the same 
in this column as in a beam fixed horizontally at both ends and 
loaded at the middle point. From the table, page 86, we have 


s 


m 


wiy 

si 


and y m = 


WP 

192El * 


, Wl 

Solving each equation with respect to y, and equating the results, 

we have 

5 l 2 

y = — tn — 

' Xw 24 Ey r 


If we replace -j y by q , we have 

_qP 
y m yt > 

fc 

which substituted in equation (390) gives 

IV' = - Sc ^ 


qF 
I + 


This is Rankine’s formula for the breaking load of a column with 
square ends. 

The suppositions made in this deduction are only approxi¬ 
mately true for columns, as may be seen by comparing the values 
of q obtained by experiment with those obtained by substituting 
the modulus of rupture of any material for s m and its coefficient of 
longitudinal elasticity for E in the equation 



Round-end Column, etc— In the Rankine formula for the 
breaking load of square-end columns, which transformed is 








COLUMNS. 


147 


, W' W'ql 2 

‘ A + Ar 2 ’ 


IF' 

the unit stress, s /, is the sum of the unit stress due to simple 

W'ql 2 

compression, and the unit stress d ue t0 flexure. The unit 

stress due to compression is assumed to be independent of the 

form of the ends; if, therefore, we represent the breaking loads of 

square, square and round, and round-end columns by W', W ", and 

# . W' W" W'" m 

the unit compressive stresses will be The 


W 


unit stress due to flexure, however, is assumed to vary for the 
same load with the form of the ends. Assuming that this vari¬ 
ation is as in Euler’s formulas, the unit stresses in flexure for 
square-end, round and square-end, and round-end columns will 
be as i :f: 2. The total unit stresses in the round and square-end, 
and round-end columns will therefore be 



W" sW"ql 2 
h + ' 2 Ar 2 ’ 


or W" = 



H 


3 # ’ 
2 r 2 


W'" 2W'"ql 2 

c r — -_i_- - — 

A ^ Ar 2 ’ 


or W'" = 


s/A 


2 ql 2 ' 


Working Formulas. 

If W' = breaking load of a square column, 

A = its area of cross-section, 
b = least dimension of cross-section, 

W' 

— = its breaking load per square inch of cross-section, 

/ = its factor of safety, 
w' = its safe load, 

— = its safe load per square inch of cross-section, we shall 
have for Gordoll’s jormulas: 


W' = 


W'A 

A 



s/A 


i + 


q/l 

b 2 


(square-end column); 



















148 


CIVIL ENGINEERING. 


W" = 


W'A 

A 



w" jA 

~aT 


s/A ] square and 
3q'l 2 1 round-end 
lJr 2b 2 column; 


w m = 


W"'A 

A 



} 

2 q'P ' 


round-end 

column. 


Rankings formulas may be derived from Gordon’s formulas 
q Q r 

by substituting for -j~ 2 . 

The factors now used with these formulas are the following; 


Wooden Columns (Gordon’s Formula). 


« 

Sc 


Yellow pine. 

5,000 


Oak. 

4,500 

< < 

White pine. 

3,5oo 

< < 


Cast-iron Columns (Gordon’s Formula). 



sS 

4' 

(circular). 

! 

q> 

(rectangular). 

Hollow cast iron. 

80,000 

7 / 

0 

tuVt 



Steel and Wrought-iron Columns (Rankine’s Formula). 



Sc' 

Q 

Medium steel. 

Soft steel. 

50,000 

45 , 00 ° 

40,000 

3 * trim 

i ( 

t ( 

Wrought iron. 


Factors of Safety.—The factors of safety employed with 
these formulas are: five for wooden columns; eight for cast-iron 
columns; four for steel and wrought-iron columns in buildings; 
five for steel and wrought-iron columns in bridges. 


f 







































COLUMNS. 


149 


By the use of either the Gordon or the Rankine formula it is 
easy to determine either W', the breaking load, or w', the safe 
W 

load, the breaking load per square inch of cross-section, or 
w' 

—, the safe load per square inch of cross-section, if all the dimen¬ 
sions of the column are given. The problem of determining 
the dimensions of cross-section corresponding to a given break¬ 
ing or safe load is a much more difficult problem. This in¬ 
volves the solution of equations of a higher degree than the 
second. 

For the above reason tables of breaking or safe loads for 
columns of different lengths and different dimensions of cross- 
section are found in all standard engineering manuals, and the 
engineer makes use of these whenever he is required to find 
the dimensions of cross-section corresponding to any given load 
and length. 

Other Formulas.—The formulas now generally applied to 
wooden columns are right-line formulas, or those containing 
only the first powers of l and b. These formulas are easy of 
application. They are usually of the form 


w' cl 


(396) 


in which w’ = total safe load of a square column; 

A = area of cross-section; 

s c = safe unit stress of material in compression; 

1 = length of column in inches; 
b = least dimension of cross-section in inches; 
c = a constant determined by experiment. 

Stanwood’s right-line formula for white-pine columns with 
square ends is of this form: 


W\ = 800 


10I 

T* 


(397) 


w r 

In this formula w x is the safe load per square inch, or j-. 

J 





CIVIL ENGINEERING. 


J 5° 


Other formulas of this class are found in engineering hand¬ 
books.* 

Eccentric Loading.—The column formulas given are based 
on the hypothesis that the load acts along the axis of the column. 
If the load is placed eccentrically, the safe load must be decreased 
so that the maximum fiber stress in the cross-section shall not 
exceed its safe value. 

Built-up Columns.—In order that there should be no waste 
of material, it is desirable that a column should be equally 
strong to resist bending in all directions. This is only true when 
the column is of circular cross-section either solid or hollow. 
If the cross-section is one of the regular polygonal forms which 
can be inscribed in a circle, as a hexagon or square, it approxi¬ 
mately fulfills the required condition. 

A built-up column is one made of two or 


Q 



as to act as a single strut. The aim in design- 


H 

Fig. 32. 


which shall act like a regular inscribed poly¬ 
gon, and be as nearly as practicable equally 
strong to resist bending in all directions. 

If a column is made up' of two rectangular 
pieces, as those shown in cross-section in Fig. 32, it will fulfill 
the required conditions as nearly as practicable if the unit stresses 
on the edges AB and AC are equal. They will be equal to each 
other if the moment of inertia or radius of gyration of the com¬ 
bined section about GH is equal to that about EF. This follows 
from the values of s c ', page 147. 

Therefore let it be required to find the distance between 
the inner faces of the rectangles AB and CD when the moments 
of inertia of the combined cross-section about EF and GH are 
equal to each other. 

Let A = area of each rectangle; 

I' = moment of inertia of the rectangle AB about EF: 


r = radius of gyration 
i'= moment of inertia 


c c 


11 


r' = radius of gyration ‘ ‘ 


I" = moment of inertia 


< c 


< < 


< < 


l ( 


l l 


l ( 


c c 


(( 


(( 


(( 


l i 


(i 


(( 


( ( 


( ( 


(< 


( ( 


< ( 


IK- 


( ( 


GH. 


* Cambria Handbook, pp. 368 , 369 . 
















COLUMNS . 


*5 1 

The moment of inertia of the combined section about EF 
is therefore 

2l' = 2Ar 2 .(39R) 

The moment of inertia of the combined section about GH may 
be determined from the following principle of mechanics: 

The moment oj inertia oj any mass with reference to any axis 
is equal to the moment oj inertia about a parallel axis through its 
center oj gravity , plus the product oj the mass into the square oj the 
distance between the two axes. 

Hence we have for the moment of inertia of the rectangle AB 
about the axis GH through the center of gravity of the combined 
section 

I"=i'+A&=Ar' 2 +A&, .... (399) 

in which k = distance between the axes IK and GH , or between 
the center of gravity of the area of the rectangle AB and the 
center of gravity of the combined section. 

For the combined section we have 

2 r' = 2Ar ' 2 + 2 Ak 2 .(400) 

By hypothesis the distance between IK and GH is such that 
2/" = 2/', hence 

2Ar 2 = 2Ar ' 2 + 2Ak 2 , .(401) 

or 

r 2 = r ' 2 + k 2 , or k = \^r 2 — r' 2 , .... (402) 

from which we can determine the value of k, when r and r' are 
known. The distance between the inner faces of the two rectan¬ 
gular pieces is 2 k — b, in which b is the breadth of the cross-section 
AB. 

If r f is so small with respect to r that its value in the second 
member may be omitted, we have 

k = r .(403) 

This value of k is slightly greater than its true value. 

In practice it is not unusual to make k = r in built-up metal 

columns. 





CIVIL ENGINEERING. 


T 52 


Application.—Let the dimensions of each of the rectangular 
beams be 16X4 inches. Then 


r 


2 



d 2 

12 


16 X16 
12 


64 

3 


21.3, 



12 


4X4 

12 



hence r 2 =r ' 2 + k 2 becomes 


2i-3 = i-3 + * 2 » 
k 2 ~ 20, 

£ = 4.47 + inches. 


In this problem if k had been made equal to r, its value 

would have been V21.3 or 4.62+; the difference between the 
two values is only 0.15 of an inch. 

Having determined the value of k we may easily find the 
distance between the inner faces of the rectangles, which may be 
represented by b', by substituting the value of k and b in the 
equation 

b' = 2k — 6 = 4.94 inches. 

In structural-metal handbooks the values of r , r', and the 
distance of the center of gravity from the inside face are given 
for all structural forms used in designing columns of wrought 
iron and steel.* The spacings for standard channels are also given 
in handbooks.! 

Column Design.—Wooden columns may be solid, and of square 
or circular cross-section, or built-up in the form of a 
box or other design; cast-iron columns are hollow and 
of circular or rectangular cross-section; wrought- 
iron and steel columns are built-up columns formed 
of plates, channels, angles, Z and I bars, so ar¬ 
ranged as to give a cross-section whose moments 
of inertia are approximately equal about bisecting 
lines parallel to different sides of the cross-section. 
A very common form is made of two channels placed 
back to back or face to face (Fig. 33), so spaced 



* Cambria Handbook, pp. 158-18^. 


t Ibid., p. 219 . 



















COLUMNS. 


J 53 


that the radius of gyration of the cross-section about A A shall be 
equal to the radius of gyration about BB. 

To find the number of square inches in the cross-section of a 
column to support any given weight, or to determine the breaking 
weight of any given column, we must substitute in the formulas 
given the values of the factors c and q, the values of A and r in 
terms of d in inches, and the value of l in inches. The resulting 
equations must then be solved either for W or d. The values of r 
in terms of d for the cross-sections employed in the design of 
wooden and cast-iron columns are: 


Rectangle, 

Circle, 

Hollow square, 
Hollow circle, 


d = shorter side, 
d — diameter, 

•> 

d = outer diameter, 

. 

d r = inner diameter, 


_ 


d 2 


12 


,2 Jl. 

i6’ 


y2 - 


, 2 _ 


d 2 + d ' 2 

d 2 + d ' 2 
16 * 


The values of r in terms of d for steel and wrought-iron 
shapes employed in constructing built-up columns will be found 
in structural-metal handbooks. To save 
the work of calculating areas of cross-sec¬ 
tion, tables are given in engineering hand¬ 
books in which the ultimate unit stresses 
are tabulated for columns of different 
materials, to correspond to all ordinary 

l l 

ratios of -j or — 
d r 

Stay-plates, etc.—At their ends, the 
channels of a steel or wrought-iron built-up 
column are fastened together by stay- or 
batten-plates, and between the stay-plates 
the channels are held together by lacing- or 
lattice-bars (Fig. 34). Lacing is formed of 




- 1! 


H 

0 ■■ 


l|0 

II 


1' 

O'l 


> 

'i 5 

a 

ll 

°'! M 

5 

J° 

• 


Jl 


Lacing 


Latticing 


Fig. 34. 


* Cambria Handbook, pp. 192-195. 


































154 


CIVIL ENGINEERING. 


single strips; latticing or double lacing is formed of double strips, 
often riveted together at their intersections as shown in the figure. 

The length of the stay-plates is usually at least equal to the 
greatest dimension of the cross-section of the column. The 
greatest distance between the rivets of the lacing or latticing 
on each channel may be determined by the proportion 

l:r'::L:r, . (404) 

in which 1 = distance between the lacing-rivets of the same channel; 
r' = least radius oj gyration of a single channel; 

L = length of the column; 

r =least radius of gyration of the column section, which 
is also the greatest radius oj gyration of a single 
channel. 

Each channel is therefore divided into short columns between 
rivets, each of which is as stiff to resist buckling as the column 
itself. In practice the distance between rivets is less than this 
theoretical distance; the angle between the lacing-strips is usually 
60 degrees, and between latticing-strips 90 degrees. The values 
of r' and r are given in structural-metal handbooks.* 

The thickness of the stay-plates and the lacing-bars may 
be computed on the theory that the column, if used as a beam, 
should be as strong to resist shearing when the channels are 
placed with the axis BB, Fig. 33, vertical as it is when placed 
with the axis BB horizontal. In the first position we may assume 
the total shear to be resisted by the lacing-strips, and in the 
second by the webs of the channels themselves. 

If we determine the greatest central load which can be borne 
by the channel webs without buckling or shearing, the stay-plates 
and lacing-bars may be proportioned to bear the same load. 
They may be computed as a truss by the methods hereafter 
described. 

Application.—Let it be required to design, with the aid of 
a structural-metal handbook, a medium-steel bridge column of 
two channels with pin-connected ends. The column is to be 20 
feet long and to safely resist a compressive force of 40,000 pounds. 

Consulting a handbook we find a table giving the strength 


* Cambria Handbook, pp. 162-165, columns 11 and 14. 





COLUMNS. 


155 


oj steel columns* * * § in which the unit breaking stress of columns 
with different ends is given for the different values of the- ratio 
of the length in feet to the least radius of gyration in inches, 
L , „ 

or —. We shall call this Table A. Before we can use this 

r 

table, therefore, we must know approximately the radius of 
gyration of the channel about the axis BB, Fig. 33. If we turn 
to the table giving the properties oj standard channels ,f which 
we shall call Table B, we find that the values of the radii of 
gyration about the axis BB , or the greatest radii of gyration, 
except for the very lightest and very heaviest channels, vary 
between 2 and 4 inches. 

For light columns we shall therefore have r= 2 inches, for 
intermediate columns r = 3 inches, and for heavy columns r = 4 
inches. Unless the character of the column is known, it is usual 
to enter Table A with the value of r for intermediate columns, 
or 3 inches. 

Since the length of our column is 20 feet, the most probable 

value of ^ is 6.7; opposite the ratio 6.7 in Table A J we find the 

ultimate unit strength of a pin-connected column, by interpola¬ 
tion, to be 36,800, and above the table we find the factor of safety 
for bridges to be 5. The allowable unit stress is therefore 
7360 pounds. Since each channel must support a weight of 
20,000 pounds, its area of cross-section should be approximately 
2.7 square inches. Turning to Table B § we find the lightest 
channel having this area to be the 7-inch, 9.75-pound channel 
whose area is 2.85 square inches and whose radius of gyration 

is 2.72 inches. The ratio ^ for this channel is 7.4 and the cor¬ 
responding allowable unit stress 6954 pounds. Its total allow¬ 
able stress is therefore 6954X2.85, or 19,819 pounds, a little less 
than 20,000. The next lightest channels are the 6-inch, 10.5- 
pound channel, whose area of cross-section is 3.09 square inches 
and whose radius of gyration is 2.21 inches; and the 8-inch, 


* Cambria Handbook, pp. 194, 195. 

f Ibid., pp. 162-164. 

X Ibid., p. 194. 

§ Ibid., p. 162. 




CIl/IL ENGINEERING. 


*5 6 

11.25-pound channel, whose area of cross-section is 3.35 square 
inches and whose radius of gyration is 3.10 inches. The value 

of ^ for the former is 9, and for the latter 6.4. The latter cor¬ 
responds more nearly to the probable value in our table, hence 
we will adopt it in our next trial. The allowable unit stress for 
this channel is 7532 pounds; hence the allowable stress of the 
entire cross-section is 3.35X7532 = 25,232 pounds, which is greater 
than 20,000 pounds. The adopted channel is therefore the 8-inch, 
11.25-pound channel. 

For the distance between the backs of the channels we may 
employ the rule heretofore deduced that the distance between the 
centers oj gravity oj the channels should be equal to twice the radius 
of gyration of each channel about the axis through its center of 
gravity perpendicular to the web. From Table B we find R' = 3.10 
inches;* * * § the distance between the centers of gravity of the channels 
should therefore be 6.20 inches. From the same table we see that 
the distance between the centre of gravity and the back of the 
channel selected is .58 inch.f Hence the distance between the 
channels back to back will be 6.20 — 1.16 inches = 5.04 inches. 
Since the width of the flange of the channel from the same table 
is 2.26 inches, the outside dimensions of the column will be 8 X9.56 
inches. The stay-plates of such a column would be made up of 
fV-inch plates; the lacing-strips would be 2 inchesX ts inch in 
cross-section, and would make an angle of 60 degrees with each 
other. They would be fastened to the channels with J-inch rivets. J 
The proper column to be employed to bear a given load may 
also be taken from tables in handbooks.? 


PROBLEMS. 

47. Find the safe load on a hollow round cast-iron column 
with square ends when the external diameter is 12 inches, thick¬ 
ness 1^ inches, length 14 feet, factor of safety 8. 

Ans. 338,956 pounds. 


* Cambria Handbook, p. 162, column 9. 

t Ibid., p. 163, column 13. 

t Ibid., pp. 250, 251. 

§ Ibid., pp. 220-282. 



COLUMNS. 


157 


48. A hollow' cast-iron column with square ends is to carry 
safely a load of 30,000 pounds; its thickness is f inch, length 12 
feet. Find the external diameter using a factor of safety of 8. 

Ans. 5.5 inches. 

49. A medium-steel column 14 feet long, with square ends, is 
made up of two 12-inch, 20.5-pound steel channels placed back to 
back. Factor of safety 4. Determine the proper spacing of the 
channels and the load that the column will carry safely. 

Ans. 145,516 pounds, and 7.82 inches. 


CHAPTER VIII. 


RIVETS AND PINS. 

Rivets. —A rivet is a short cylinder of malleable metal, usually 
headed at one end, which is employed to fasten together the 
wrought-iron or steel pieces in an engineering structure. (Fig. 35, 

_ 

A ‘ \J KJ 

. . r\ r\ , 

r\ r\. 

c 1 cj ' w 

Fig. 35. 

E and F.) The rivet E has a jull head , and F a countersunk head 
on the right end, made by upsetting the headless end. 

The ordinary sizes of rivets vary from f to 1 inch in diam¬ 
eter by differences of J of an inch; intermediate sizes are also 
made.* The f-, f-, and f-inch rivets are the ones commonly 
used. 

To make a riveted joint, holes are punched or drilled in the 
pieces to be fastened together, and the rivets, after being passed 
through the holes, are secured in place by upsetting the headless 
ends by successive blows of a hammer or by pressure. The head¬ 
less ends of rivets over one-half inch in diameter must be heated 
to a red heat before using; the upsetting produces a second head, 
hemispherical or conical, similar to the first. The holes for rivets 
are usually punched, as drilling is more expensive. In plates | 
of an inch in thickness, or less, the holes are punched to full size; 
in thicker plates the holes are punched J of an inch less than the 
full size and then reamed to the full size. The reaming removes 
the metal immediately about the hole which has been injured in 



* Cambria Handbook, pp. 334, 335. 


158 































RIVETS AND PINS. 


*59 


punching. In very important work the holes in the thin plates 
are punched and reamed, and those in plates over J inch in 
thickness are drilled. The diameter of the rivet-hole is usually 
made ^ inch greater than that of the rivet, but in computing the 
net area of the piece at a rivet-hole it is assumed as J inch greater 
than the rivet. 

If the holes for the rivets have been carefully laid out, the holes 
should line up accurately when the pieces are superposed. In 
work carelessly done all the holes do not line up accurately, and 
a conical steel pin, called a drijt-pin , is often used to bring the 
holes into line. Drifting is prohibited in all specifications for 
high-grade work. 

Riveted Joints.—When pieces are fastened together at or near 
their ends, the joint is called a lap-joint if the pieces overlap (Fig. 
35, A and D), and a butt-joint if the ends abut against each other 
(Fig. 35, B and C). A cover-strip, or fish-plate, is a third piece 
which is riveted to both pieces to be joined (Fig. 35, A, B, and C). 
A joint is single-riveted when each piece is fastened by a single 


_rs _q _o_ cs _q _r\_ 

a rz—;~r .'.. - - 1 ——- ; —zb }) 

-O-o-O-O-CT"* 




Fig. 36. 


row of rivets perpendicular to its axis (Fig. 35, A and D). It is 
double-riveted when each piece is fastened by two rows of rivets 
(Fig. 35, B and C). It is chain-riveted when it is fastened by more 
than two rows of rivets (Fig. 36). The rivets are staggered when 
they are placed in quincunx order (Fig. 3b)* The pitch or spacing 

















i6o 


CII/IL ENGINEERING. 


of the rivets is the distance between the centers of consecutive 
rivets. 

Riveted Tension-joints.—A riveted tension-joint may rupture 
by the yielding of the rivets, by the yielding of the pieces joined, 
or by the yielding of the cover-strips. 

I. The rivets may yield by shearing off at one or more cross- 
sections oj the rivets themselves. 

If each rivet which yields shears off along a single plane of 
cross-section, as in Fig. 35, B and D, the rivets are said to be in 
single shear; if each rivet can yield only by shearing off along 
two planes of cross-section simultaneously, as in Fig. 35, C, the 
rivets are said to be in double shear. 

To resist rupture by single shear we must have the total 
resistance of all the rivets in single shear equal to the force to be 
transmitted, or 

ns ^ nd^ 

— ——=0.7854 s 8 "nd 2 =F f .... (405) 

in which s 8 " = allowable unit shearing stress of the rivets in single 

shear; 

71 = least number of rivets in any of the pieces to be 
joined; 

d = diameter of each rivet in inches; 

F = total tensile force to be transmitted through the 
joint in pounds. 

From this formula we may determine the diameter of a single 
rivet when the number of rivets in the joint is assumed or known. 

Formula (405) may be written in the form 


F 


F 


71 = 


0.7854 s/'d 2 single-shear value of one rivet 3 


(406) 


in which 0.7854^%/' is the single-shear value oj one rivet whose 
diameter is d. The single-shear values of rivets of different diam¬ 
eters and for different values of s s " are given in handbooks on 
structural metal.* 

To find the number of rivets of a given size required to trans- 


* Cambria Handbook, p. 30S, third column. 







RIVETS AND PINS. 


161 


mit a given tensile stress, we need only take the single-shear value 
of a rivet from the table and substitute it with F in the formula 
(406) and deduce the value of n. 

To resist rupture in double shear the formulas become 

1.570 8 s s "nd 2 = F, .(407) 

F _ F _ 

° r 71 1.570 &s s ''d 2 double-shear value of one rivet’ 

in which i.57o8s 8 "d 2 is the double-shear value oj one rivet whose 
diameter is d. The double-shear values of rivets are also 
tabulated in the handbooks.* 

II. The plate may yield by the rivets crushing into the sides 
oj the holes. 

The plate is then said to be deficient in bearing area. The 
area of material crushed by each rivet is dt, in which d is the 
diameter of the rivet, and t the thickness of the plate. To resist 
crushing, therefore, we must have the total bearing resistance of 
the plate at the rivet-holes equal to the force to be transmitted, or 

s b "ndt = F ,.(409) 

in which s b " = allowable unit stress of the material in bearing; 
n = number of rivets; 
d = diameter of the rivets in inches; 
t = thickness of the plate in inches; 

F = total tensile stress to be transmitted in pounds. 

From this formula we may determine the thickness of the 
plate when the number of rivets and the diameter of each rivet 
are known, or we may determine either the number of rivets of 
an assumed diameter, or the diameter of an assumed number of 
rivets, corresponding to any given thickness of plate. Formula 
(409) may be written in the form 

_ F __ F _ 

n ~ s b 'dt~ bearing value of the plate and a single rivet’ 


* Cambria Handbook, p. 308, fourth column. 










162 


CIVIL ENGINEERING. 


in which s b 'dt is the bearing value of the plate, also called the 
bearing value of the rivet. The bearing values of plates of 
different thicknesses when subjected to pressure of rivets of 
different diameters and for different values of s b " are given in 
structural-metal handbooks.* 

To find the number of rivets of a given size required to trans¬ 
mit a given tensile stress we need only take from the table the 
bearing value corresponding to the rivet and plate and substi¬ 
tute it with F in the formula (410). 

If the rivets are in single shear, the value of n must be deter¬ 
mined from equation (406) if the single-shear value of the rivet 
is less than its bearing value; if the latter is less, the value of n 
is determined from equation (410). If the rivets are in double 
shear, the value of n is determined from equation (408) if the 
double-shear value of a rivet is less than its bearing value, 
and from equation (410) if the bearing value is less. In the tables 
of bearing values of rivets, the limits within which the bearing 
values are less than the single- and double-shear values are indi¬ 
cated by horizontal lines. 

III. One oj the pieces or the cover-strips may yield by tearing 
apart along one oj the transverse lines oj rivets. 

In a single-riveted joint , to resist this mode of yielding the 
resistance of the net section of each piece or of the cover-strips 
must be equal to the force transmitted, or 

s e "(bt — vdt)=F, .(411) 

in which s e " = allowable unit stress in tension or elongation; 
b = breadth of each piece in inches; 
t = thickness of the weakest piece, or of the combined 
thickness of the cover-strips, in inches; 
v = numerical coefficient, which, for a single-riveted 
joint, is equal to the number of rivets; 
d = diameter of each rivet in inches plus J inch; 

F = total stress transmitted in pounds; 

(bt—vdt) = net section of the piece or of the combined cover- 
strips at their weakest section; 


* Cambria Handbook, pp. 308, 309. 





RIVETS AND PINS . 


163 


bt—gross section of the piece or of the combined cover- 
strips at their weakest section. 

When the joint is a double- or chain-riveted one, the location 
of the weakest section and the value of v in the first member of 
the above formula will depend upon the arrangement of the 
rivets. 

In the butt-joint Fig. 36, A, the total area bt of the piece 
at the cross-section d is decreased by the area dt , or the meridian 
section of one rivet-hole; at the section e the total area is decreased 
by the area 2 dt, or the area of the meridian sections of two rivet- 
holes; but the piece cannot rupture along this section unless the 
rivet in the section at d first fails; at the section / the total area 
of the piece is reduced by 3 dt , but the piece cannot rupture at 
this section without the three rivets in the sections d and e first 
failing. I / the resistance to shearing or bearing 0] a single rivet is 
equal to or greater than the tensile resistance 0] the area dt 0] the 
piect, it is evident that the rivets may always be arranged (as in 
Fig. 36, A) to make the minimum strength of a piece at a joint 
equal to the tensile strength of its total cross-section minus tensile 
strength of the meridian section of one rivet hole, or ( b — d)t . 

In the joint Fig. 36, A, if the cover-plates are made of the 
same material as the pieces themselves, the combined thick¬ 
ness of the cover-plates must be greater than the thickness of 
either piece, since the section of weakness of the cover-plates 
is the cross-section /, which has been reduced by three rivet- 
holes, while the weakest section of each piece has been practi¬ 
cally weakened by only one. 

In the joint Fig. 36, B, the plane of weakness in both pieces 
and cover-plate passes through a single rivet-hole, but the cover- 
plate is a very long one. The arrangement of rivets is usually 
that shown in Fig. 36, A or C. 

Values of dt corresponding to different values of d and t are 
tabulated in structural-metal handbooks. By the use of these 
tables the value of vdt can be readily determined.* 

Pitch and Size of Rivets.—In applying the formulas under I, 
II, and III, it is necessary to assume either the size or the number 
of the rivets. In the direction of the stress the pitch of the rivets 


* Cambria Handbook, pp. 312, 313. 





364 


CIVIL ENGINEERING. 


should not exceed either 6 inches, or sixteen times the thickness 
of the thinnest plate; in a perpendicular direction it should not 
exceed thirty-two times the thickness of the thinnest plate. The 
minimum pitch oj rivets in any direction is three diameters. The 
distance between the center of the outside rivet and the edge of a 
plate should be not less than ij diameters nor more than eight 
times the thickness of the plate. 

The maximum spacing of rivets in the flanges of girders is 6 
inches; in pin-plates at the end of columns, girders, etc., it is 
usually four diameters. The diameter of a rivet-hole should never 
be less than the thickness of the thickest plate which is to be fast¬ 
ened by it; in thin plates it should usually be at least one-third 
to one-half greater. In structural-metal handbooks tables are 
given showing the size of rivets which should be used with differ¬ 
ent I beams, channels, angles, and plates,* and also rivet-spacing.f 

Compression-joints.—The only difference between the design 
of riveted compression- and tension-joints is that in the former no 
allowance need be made for the rivet-holes in determining the 
strength of the compression-pieces or cover-plates at the joints; the 
rivets are assumed to fit the holes perfectly, and to replace the 
material of the hole. In ordinary riveted compression-joints the 
rivets themselves are designed to transmit the entire stress, and 
no reliance is placed on the abutting surfaces to transmit it. This, 
however, is not true of the joints in the compression-chords of a 
bridge-truss; the surfaces in these joints may be so accurately 
planed and fitted that the stress need not all be transmitted 
through the rivets. 

Application. 

Let it be required to design a riveted tension butt-joint of 
structural steel which will safely transmit a tensile force of 60,000 
pounds. 

Then F = 60,000 pounds. 

Let s/' = 15,000 pounds = safe unit stress in tension for main 
and cover plates; 

s" =s s " = 12,000 pounds = safe unit stress in bearing and double 
shear for rivets; 


* Cambria Handbook, p. 312. 
f Ibid., pp. 313, 314. 




RIVETS AND PINS. 


165 


d = | inch = diameter of each rivet; 
t = h inch = thickness of ties. 

In a butt-joint there are usually two cover-plates, hence the 
rivets will be in double shear. 

Number of Rivets.—In the handbook the double-shear value 
of a f-inch rivet is given as 5301 pounds, and the bearing value of 
4 f-inch rivet in a ^-inch plate as only 4500 pounds.* The 
latter must therefore determine the number of rivets. 

Assuming equation (410) we have 


n= 


60,000 

4500 




Fourteen rivets will therefore be required to fasten each main 
plate to the cover-plates. 

Width of Main Plates.—If there were no rivet-holes in the 
plates which are united, the width of each would be given by the 
formula 

60,000 

bt = -=4 square inches, whence b = S inches. 

15,000 


We will assume the rivets to be placed in rows across the plate, 
four in each row, except the outside rows, which contain but two. 

The gross width of the plates through the outside rows must 
therefore be 8 +2(f) inches = 9! inches. 

The gross width through the next row is deduced from the 
requirement that the tensile strength of the plate, minus four rivet- 
holes, plus the strength of two rivets in bearing must be equaL 
to the force transmitted, or 

1 

15,000 X \ X [b — 4(1)] + 2 (4500) = 60,000, 
or £ = 10.3 inches. 

This is therefore the weakest section of the plate, as each 
section through rows nearer the middle of the joint has the same 
reduction in area, but is strengthened by the resistance of a 
greater number of rivets. 

O 


* Cambria Handbook, pp. 308, 309. 





i66 


CIVIL ENGINEERING. 


It is necessary to ascertain whether this width allows proper 
rivet-spacing. The rivets in each row may be placed at intervals 
of 2\ inches and an interval of i J inches must be left between the 
outside rivet and the edge of the plate; as this spacing requires 
a plate only io inches wide, the main plates are sufficiently wide. 

Thickness of Cover-plates.—The weakest section of the cover- 
plates is through the rows of rivets on either side of the center of 
the joints, and hence at a point where the cover-plates are reduced 
by four rivet-holes. 

The combined thickness of the cover-plates is therefore de¬ 
duced from formula (411) by making v = \ and substituting the 
proper values for the other quantities in the formula, or 

60,000 

t= •-7 -7=0.50. 

15,000(10.3-3.5) 

Each plate should therefore be 0.30 inch thick. The standard 
plate nearest in thickness to this is A inch thick. 

Since the combined thickness of the two cover-plates is more 
than the thickness of each tie, it is unnecessary, in this case, to 
test the cover-plates for bearing, since their bearing value for each 
rivet must be greater than the bearing value of either tie. 


PROBLEM. 


50. A double-cover butt-joint unites two 8 Xf" plates. Each 
fish-plate is 8Xf". Eleven f-inch rivets unite each tension- and 
fish-plate; they are arranged in six rows, which from the middle of 
the joint contain respectively two, three, three, two, and one rivet. 
What is the maximum stress which can be transmitted through 
the joint? Stresses as on page 164. A ns. 58,311 pounds. 


Pins and Pin-joints.—Whenever three or 
more pieces of an engineering structure meet 
at a common vertex, and the stresses in the 
pieces are longitudinal stresses in equilibrium, 
a pin-joint may be employed. (Fig. 37.) 

The tensile stresses are usually carried by 
eve-bars, and the compressive stresses by 
columns; each piece is provided with a pin¬ 
hole at its extremity to receive the pin. 









RIVETS AND PINS. 


167 


Eye-bars.—An eye-bar is a wrought-iron or steel bar of cir¬ 
cular or rectangular section, having at one or both extremities an 
enlarged, flattened head, with a circular hole or eye to receive the 
pin. (Fig. 37.) The axis of the pin-hole intersects the axis of 
the bar, and the hole is about ^ of an inch larger than the diam¬ 
eter of the pin. Eye-bars are usually made by upsetting the ends 
of straight bars and forging the ends into shape by hydraulic 
pressure in suitable dies. The form and thickness of the metal 
about the pinhole is so designed that the weakest cross-section in 
an eye-bar shall be in the straight portion of the bar. This is 
effected in a bar of uniform width by proportioning the depth of 
the eye-bar as follows: (Fig. 37.) 

Depth of bar. Radius of Hole. Radii of ABC and DEF. 

1 From | to 2J 2J to 3J 

Standard eye-bars vary in width from 2 to 10 inches, and are 
made to fit pins varying in diameter from if to 10 inches. (See 
tables in structural-metal handbooks.*) 

Columns.—Pin-connected steel or wrought-iron columns are 
usually made in the form of two channels laced together; the 
pin-hole is made through the webs of the channels, or through 
the stay-plate. The metal about the pin-hole is reinforced by 
additional plates riveted on to prevent the pin from crushing into 
the web. The thickness of the reinforced area of the web is 
determined by the condition that the safe bearing resistance of 
the reinforced web shall be at least equal to the force transmitted 
through the columns, or 

F 

ts b d — F, . . i = J77~^ .(4^2) 

in which t = total thickness of a single web and its reinforcing-plate 
in inches; 

F = total compressive stress in pounds transmitted through 
a single web; 

s b " = allowable unit stress of material in bearing; 
d = diameter of the pin in inches; 
ts h "d = bearing value of the pin on the reinforced web. 


* Cambria Handbook, pp. 330, 331, 337 - 339 - 





i68 


CIVIL ENGINEERING . 


The bearing values of pins of various diameters in one-inch 
plates are tabulated in handbooks of structural metals for conve¬ 
nience in computation.* 

Pins.—Pins are circular forged steel cylinders having either a 
head and a nut or two nuts at their extremities. The standard 
sizes vary in diameter from i to io inches (see tables in structural- 
metal handbooks).! 

A pin is simply a beam acted upon by a system of non-coplanar 
bending forces in equilibrium. The pin must be designed to resist 
the maximum bending moment and the maximum shear to which 
it is subjected. As the forces acting on a pin do not, as in the 
beams heretofore considered, all act in the same plane containing 
the axis of the pin, to determine the maximum bending moment 
and maximum shear each force is resolved into its vertical and 
horizontal components, and hence the entire system into two co- 
planar systems, one in the vertical and the other in the horizontal 
plane through the axis of the pin. 

Let M = resultant moment at any section in inch-pounds; 

Mh= moment at the same section due to the horizontal 
components in (inch-pounds); 

M v = moment at the same section due to the vertical com¬ 
ponents in (inch-pounds); 

S s = resultant shear at any section in pounds; 

S h = shear at the section due to the horizontal components 
in pounds; 

S v = shear at the same section due to the vertical com¬ 
ponents in pounds. 

Then at every section of the beam we shall have 

M — ^Mj ‘ 2 + M v 2 (413a) 

and 

S s = VSh 2 +S V 2 .(4136) 

The dangerous section of the pin will be when M is a maximum. 
The pin may be tested for shearing strength at the section where 
S s is a maximum, but this is usually unnecessary. 


* Cambria Handbook, p. 315. 
t Ibid., pp. 336, 337. 







RIVETS AND PINS. 


169 

The bearing strength of the ties and struts which abut on the 
pin should always be tested, as they usually determine its size. 

The location of the dangerous section along the axis of the 
pin may usually be ascertained by constructing and inspecting 
the curves of bending moments for the vertical and horizontal 
components separately. 

The location of the section of greatest shear may usually be 
ascertained by constructing and inspecting the lines of shear for 
the vertical and horizontal components separately. 

Application.—Assume five forces shown in Fig. 37 as inter¬ 
secting at a common point A. Let AB = 90,000 pounds, AC = 
30,000 pounds, AD = 150,000 pounds, A £ = 30,000 pounds, and 
AF = 84,850 pounds. Substitute for the force AB two equal 
components, and let each be transmitted through an eye-bar, of 
the form shown in Fig. 38, to a pin whose axis is perpendicular 



Fig. 38. 

to the plane of the paper at A, Fig. 37. In a similar manner let 
AD and AF be transmitted through two eye-bars, so arranged that 
the force transmitted through each bar shall be the same, and 
the resultants of each pair of forces shall act through the middle 
point of the axis of the pin. The angle EAF = 45 degrees. 

Let the force AE be transmitted through a column made 
up of two laced channels, as in Fig. 38, also symmetrically dis¬ 
posed with respect to the middle point of the pin. Assume 
the distance between the channels to be 3J inches, and the flange 









































































































































































































































CIVIL ENGINEERING. 


170 

of each channel to be 2 J inches wide, the web to be £ inch thick, 
with a reinforcing-plate £ inch thick. 

Let the force AC be transmitted through a rectangular plate J 
inch thick, as in Fig. 38, whose axis passes through the middle 
point of the pin. We shall then have a pin-connected joint in 
which the pin has neither motion of translation nor rotation. 
Such a joint is shown in plan in Fig. 38. 

Maximum Shearing and Bending Moments.—From Figs. 
37 and 38 we may construct the table given below. In the first 
column we write the pieces in- their order beginning with the 
outside of the pin. The second column is obtained by dividing 
the total stress in the piece by the allowable unit stress in ten¬ 
sion, assumed as 15,000 pounds; in the third column are stand¬ 
ard eye-bars taken from the table which have approximately the 
dimensions of cross-section required; the fourth column gives 
the total stress in each piece; the fifth column gives the hori¬ 
zontal components of these stresses; the sixth column gives the 
shears in the pin due to the horizontal components; the seventh 
and eighth columns give the vertical components and vertical 
stress; and the ninth column gives the maximum shearing 
stresses in the pin, obtained by solving the equation s s = \ // S; i 2 -hs v 2 . 


I 

Piece 

2 

Area 

Re¬ 

quired 

3 

Eye- 

bars 

Selec¬ 

ted. 

4 

Stress¬ 

es. 

s 

Horizon¬ 
tal Com¬ 
ponents. 

6 

Horizontal 

Shear. 

7 

Vertical 

Com¬ 

ponents. 

8 

Vertical 

Shears. 


Sq. In. 

Ins. 

Pounds. 

Pounds. 


Pounds. 

Pounds. 

Pounds. 

AB 

3 

4 X 1 

45,ooo 

+ 45,000 

1 










45,ooo 



AD 

5 

5 X 1 

75,ooo 

-75,ooo 

J 

' 




AE 



15,000 



• 30,000 

-15,000 








1 

’ 15,000 

AF 

2.83 

4x1 

42,425 

+ 30,000 

a 


+ 30,000 

J 

1 











15,000 

AC 

2.0 


30,000 



\ 0 

— 30,000 

J 

1 











15,000 

AF 

2.83 

4x1 

42,425 

+ 30,000 

a 

*N 


+ 30,000 

J 

1 











f 15,000 

AE 



15,000 



. 30,000 

-15,000 

J 


AD 

5 

5 X 1 

75 ,ooo 

-75,000 

J 

'I 



• 







l 45,ooo 



AB 

3 

4 X} 

45 , 00 ° 

+ 45,000 

J 





9 

Resultant 

Shear. 


Pounds. 
45,000 
30,000 

} 33,540 
15,000 
15,000 

} 33,540 
30,000 
45,ooo 
































RIVETS AND PINS. 


17 I 

Forces acting to the right and those acting upwards are con¬ 
sidered positive , those acting in a contrary direction are considered 
negative. 

We may also write out a second table to determine the maxi¬ 
mum bending moment. In the table given below, the first 
column gives the pieces in their order; the second gives the total 
stresses; the third gives the horizontal component of the stresses’; 
the fourth gives the distances between the axes of the pieces; 
the fifth gives the moments of these horizontal components about 
points on the axis of the pin; the sixth gives the vertical com¬ 
ponent of the stresses; the seventh gives the distance between the 
axes of the pieces; the eighth gives the moments of the vertical 
components; the ninth gives the maximum bending moment 
derived from the solution of the equation 

M=VmJ+m}. 


Piece 


AB 

AD 

AE 

AF 

AC 


btresses. 


Pounds. 

45,000 

75,000 

15,000 

42,425 

30,000 


Horizon¬ 
tal Com¬ 
ponents. 


Pounds. 
+ 45,000 

-75,000 


+ 30,000 


Be¬ 

tween 

Bars. 


Inches. 


f 


Moments. 


(Inch-lbs.) 

o 

39.375 
35,625 

54.375 
54,375 


Vertical 

Compo¬ 

nents. 


Pounds. 


-15,000 
+ 30,000 
— 30,000 


Be¬ 

tween 

Bars. 


Inches. 


3 

¥ 


Moments. 


(Inch-lbs.) 

O 

O 

o 

9,375 

13,125 


Resultant 
Moment , 


(Inch-lbs.) 

39,375 

35,625 

55,177 

55,936 


In forming this table the stress of each piece is assumed to 
act along its axis. A simpler method of determining the bend¬ 
ing stresses in pins will be given hereafter under Graphic Statics. 

In packing the pin, or in arranging the order of the pieces 
on it, it will be observed that the smaller the maximum stress 
in any of the pieces, the smaller will be the maximum shear and 
the maximum bending moment; hence if the required pin diam¬ 
eter is too large, it may be decreased by substituting two or more 
bars for each of the two single bars carrying the greatest stress. 
It will be observed that the maximum shear and the maximum 
bending moment are diminished by alternating positive and 

























172 


CIVIL ENGINEERING. 


negative bars, and by placing the bar with the greatest negative 
stress adjacent to the bar with the greatest positive stress. The 
single bar is usually the tension-bar having the smallest stress. 

If the bars are separated by any considerable distance, 
washers are used between them; if no washers are used, an allow¬ 
ance of re of an inch is made for each space between bars, in 
determining the length of the pin. The flanges of the channels 
are removed in the vicinity of the pin, when by so doing the 
maximum bending moment may be greatly decreased. In the 
example given the flanges have been left on for clearness, and 
the distance between the channels of the column has been 
reduced to make Fig. 38 small. 

Design.—Having determined the maximum shear and the 
maximum bending moment in the pin, its cross-section is designed 
by employing the usual formulas. 

For Shear.— 



4 



(414) 


in which s 8 " = allowable unit stress in shear =10,000 pounds; 
d = diameter of pin in inches; 

F = maximum shearing force = 45,000 pounds. 


Substituting and solving for d we have d= 2.4+ inches. 

For Flexure.— 


s^I 

y 



(415) 


in which s" = allowable stress in flexure = 20,000 pounds; 

Tld* 

/ = moment of inertia = — = 0.049#; 

, d 

M = maximum bending moment = 55,936 (inch-pounds). 


Substituting and solving for d, we have ^ = 3.06 inches. 

The required diameter of a pin to resist a given bending 
moment may also be taken directly from the tables in hand¬ 
books.* 


* Cambria Handbook, pp. 310, 311. 








RIVETS AND PINS. 


173 


From the table of eye-bars we see that the diameters of the 
largest pins corresponding to the bars selected vary from 4J to 6J 
inches.* 

In this example the 3f-inch pin is sufficiently large to resist 
the maximum bending moment, but is not large enough for 
bearing on any of the bars. 

For Bearing.—The bar requiring the largest bearing area 
and also the largest pin is the bar AD, which transmits a tensile 
force of 75,000 pounds. The bearing area required by AD 
may be found from the formula 


td = 



(416) 


in which / = thickness of bar in inches =1 inch; 
d = diameter of pin in inches; 
s b " = allowable unit stress in bearing in pounds = 15,000; 

F = force transmitted through bar in pounds = 75,000. 
Solving the equation with respect to d, we find that the diameter 
of the pin must be 5 inches. 

Bearing Value of Reinforced Web.—Having determined the 
diameter of the pin, the next step is to ascertain whether the 
webs of the struts AE have sufficient bearing area without a 
reinforcing-plate. 

In the formula 


td = 


F 


tn 


(417) 


making l = } inch, 
d = 5 inches, 
s b " = 15,000 pounds, 

the resulting value of F is 18,750 pounds. As each strut trans¬ 
mits a force of 15,000 pounds only, a reinforcing-plate will not 
actually be required at the pin-hole, but would ordinarily be used 
to strengthen the web, as shown in the figure. 

The bearing value of a 5-inch pin in a i-inch plate might also 
have been taken directly from the handbook,! and the bearing 
value of a J-inch plate found by dividing the value given by four. 


* Cambria Handbook, p. 331. 
t Ibid., p. 315. 







CHAPTER IX. 


SOLID BUILT BEAMS, I BEAMS, AND PLATE GIRDERS. 

Solid Built Beams.—A simple beam is a beam cut out of a single 
piece of timber. A solid built beam is a beam made of two or 
more simple beams, so fastened together as to act as a single 
beam. If the. two pieces are of equal length, they may be bolted 
together as shown in Fig. 39. 

The strength of a rectangular beam, or its power to resist 
additional loads, is given by the equation 

i_ bd 2 

s~6mWl’ .^ 4l8 ) 


in which m = a coefficient depending on the manner of loading and 

supporting the beam; 
b = breadth of cross-section in inches; 
d = depth of cross-section in inches; 

/ = length of beam in inches; 

W = total load in pounds; 

s = maximum unit fiber stress in the beam in pounds. 
From this equation it appears that, all other conditions being 
the same, the strength of a rectangular beam varies with the first 
power oj the breadth and the square oj the depth. 

If, therefore, two beams of equal breadth 
and depth are placed side by side, as shown in 
the upper drawing, the strength of the two 
beams will be twice the strength of a single 
beam. 

This will be equally true whether the beams 
are simply laid side by side, or fastened to¬ 
gether as shown in Fig. 39. In the latter case 
there will be no stress in the bolts, since their 
axes lie in the neutral surface or in a surface parallel to it. 

U4 



Fig. 39 - 









SOLID BUILT BEAMS , I BEAMS , PLATE GIRDERS. 175 


If the beams are superposed, as shown in the lower drawing, 
but not jastened together , neglecting friction, the strength of the 
two beams will also be twice the strength of a single beam, since 
the beams will act independently, each having its own neutral 
surface and the same stress in the fibers at its upper and lower 
surfaces. If the beam rests on two supports, under the action of 
the external forces the lower surface of the upper beam is length¬ 
ened, and the upper surface of the lower beam is compressed; 
hence these surfaces slide along each other. If, however, the 
beams are bolted together as shown in Fig. 39, and this sliding 
is prevented, the neutral surface of the solid built beam will be at 
the surface of contact of the two beams, and the strength of the 
solid built beam will be jour times that of a single beam. The 
upper beam will now be wholly in compression, the lower one 
wholly in tension, and the bolts will be subjected to a horizontal 
shearing stress due to the tendency of the beams to slide upon each 
other. 

Rolled I Beams.—A rolled I beam is a solid metal beam of 
steel or wrought iron, whose uniform cross-section is of the form 
of the letter I. The upper and lower horizontal branches of the 
I are called the flanges; the vertical connecting part is called the 
web. 

Rolled I beams are made of standard sizes whose dimensions 
may be found in structural-metal handbooks.* To determine the 
load which may be placed on such a beam we utilize the formula 


s"I 2 s"Ar 2 
Mfn = / d } 


(419) 


in which M m = bending moment at the dangerous section in terms 

of the load in pounds and the length of the beam 
in inches; 

5" = safe unit stress of material in pounds; 

— section modulus of beam, in which dimensions are 

/ 

in inches; 

A =area of cross-section in square inches; 


* Cambria Handbook, pp. 158-161. 







176 


CiriL ENGINEERING. 


7 =radius of gyration of cross-section about the neutral 
axis in inches; 

d = depth of the beam in inches. 

Application.—What weight may be safely distributed uni¬ 
formly along an 8-inch 18-pound I beam, 12 feet long, if 5 = 
15,000 pounds? 

From the table, page 86, we have M m = \Wl\ from the hand¬ 
book -y = i4.2. Substituting these values in (419) we have 

y 

JITX12X12 = 15,000X14-2, 

whence W= 11,833 pounds. This includes the weight of the 
beam, 12X18 pounds, or 216 pounds. 

Plate Girder.—A plate girder, or built-up I beam, is a steel 
I beam in which the flanges and web are separate pieces fastened 

together by rivets, as shown in 
Fig. 40. The web is a metal plate 
whose thickness is usually from 
| to J inch, and whose depth 
varies from J to T V the span, in 
heavily loaded girders, and from 
° it to -g-V the span, in light ones. 
Each flange consists of two 
angles which are riveted to the 
upper and lower edges of the web as shown. The angles are 
made with equal or unequal legs, varying in width from 2 to 7 
inches, and in thickness from J to -J inch. The area of each 
flange may be increased by one or more flange-plates riveted to 
the horizontal legs of the angles. The area of cross-section of 
the flange-plates should not exceed that of the angles unless the 
maximum size angles are employed. The flange-plates are in 
contact with the angles only and not with the edges of the web, 
which are always kept inside the faces of the horizontal legs of 
• the angles. 

At the supports and at points where the girder receives con¬ 
centrated loads the web is stiffened by angles or tees which are 
riveted to it in pairs on its opposite sides. The ends of these 
stiffeners abut on the flanges. 


D 












SOLID BUILT BEAMS, I BEAMS, AND PLATE GIRDERS. 177 


The different parts of the girder are usually fastened together 
by J- or J-inch rivets; the larger size is used in heavy girders, and 
the smaller in light ones. The load of a girder may be placed 
on the upper flange, or upon brackets attached to the web. 

In determining the stresses in a plate girder one of three 
methods may be utilized: 

1. The girder may be treated as a solid beam. That is, the 
longitudinal stresses may be assumed to vary from the neutral 
axis to the surface, and the vertical shear to be uniformly dis¬ 
tributed over the entire area of cross-section. 

2. The flanges of the girder may be designed to resist the 
longitudinal stresses without the assistance of the web, and the 
web may be designed to resist the vertical shear without the 
assistance of the flanges. This method gives the greatest factor 
of safety and is the one usually employed. 

3. The girder may be so designed that the longitudinal stresses 
are distributed over the flanges and web, approximately as in 
the first method, but the vertical shear is resisted by the web 
alone. This method is also employed in designing girders. 

First Method.—The first method cannot well be employed 
in designing girders, but it may be employed in determining 
the load which may be placed on a girder whose section modulus 
is known. If the girder is of uniform section, we may determine 
M in terms of W from page 86 and substitute it in the formula 


M 


m 


S"I 


( 419 ) 


in which M = the bending moment at the dangerous section in 

(inch-pounds); 

s" = allowable unit stress in pounds; 

-— = the section modulus of the beam in inches. 

y 

Second Method.—In the second method the girder is designed 
as follows: The depth of the web, which is practically that of 
the girder, is first assumed at } to the span, depending on 
the load and the allowable depth. A girder is often placed 
in a position in which the most economical depth is not 
allowable. 




CIVIL ENGINEERING. 


178 


The Flanges.—Since the flanges are designed to resist the 
entire longitudinal stresses, the moments of these stresses about 
the neutral axis, EE , Fig. 40, must, at every section, be equal to 
the bending moment. These stresses are assumed to be uniformly 
distributed over the area of the flanges, hence the center of stress 
will be at the center of gravity of each flange. The flange is 
therefore so designed that at every cross-section 


s"A r d\=or >M, .(420) 

in which s" = safe unit longitudinal stress in pounds; 

A' = net area of cross-section of each flange in square 
inches; 

d 1 = distance between centers of gravity of flanges in 
inches; (This is called the effective depth of the 
girder.) 

M = the bending moment at the cross-section in (inch- 
pounds). 

From this formula we can determine A' at any section when 
we know the values of s", d\, and M. The first and last can be 
readily determined, but the second must be assumed. In the 
tables in structural-metal handbooks are given the properties of 
the angles employed in making up plate girders.* From these 
tables it will be seen that the distance from the center of gravity 
of an angle to its back varies only from 0.60 to 2.41 inches for angles 
employed in plate-girder construction; therefore if d\ is assumed 
from o to 2 inches less than the depth of the web, it will be 
sufficiently close to its true value to select angles for trial. Its 
true value, which must not be less than its assumed value, may 
be determined when the composition of the flange is definitely 
fixed or the angles and flange-plates are definitely selected. 

Having determined the true net area , the true gross area of 
a flange is found by adding to the net area the area of the 
rivet-holes in a vertical plane through the center of a rivet-hole. 
The area added for this purpose in Fig. 40 is the area of cross- 
section of one of the web-rivet holes, . 4 . 4 , plus the area of cross- 
section of the two flange-rivet holes in the same cross-section. If 


* Cambria Handbook, pp. 166-179. 






SOLID BUILT BEAMS , / BEAMS, AND PLATE GIRDERS. 179 

the web- and flange-rivet holes are not in the same plane, they 
are nevertheless so considered unless the oblique plane through 
them has a net area 30 per cent greater than the net area of the 
vertical section. 

The gross area of the compression-flange is made equal to that 
of the tension-flange, although this is not really essential if the 
rivets fully fill their holes. 

The Web.—The web of a girder is usually made of plates * of 
the same depth and thickness, spliced end to end. Sometimes, 
when the load is heavy and the girder is long, the plates vary in 
thickness. 

The web may yield because of insufficient bearing area at the 
flange- or stiffener-rivets, or because of insufficient thickness to 
resist the vertical shear or to prevent buckling. If at the point 
of maximum vertical shear, usually the supports, the web is thick 
enough to give sufficient bearing area on the stiffener-rivets, and 
thick enough to resist the vertical shear, buckling may be pre¬ 
vented by the use of stiffeners, and the yielding of the flange- 
rivets by increasing their number. 

The thickness to resist the bearing of the stiffener-rivets is 
determined from the formula 


t = 


(V.Y 


bearing value of one rivet X nd ’ 


(421) 


in which n = number of rivets in the stiffener; (The number of 

these rivets must be sufficient to resist the bearing 
and double shear of (F«) m .) 
d — diameter of rivets in inches; 

/ = thickness of web in inches; 

m = vertical shear at support or the maximum vertical 
shear in pounds. 

The thickness of the web to resist shearing is determined from 
the formula 


( V.)„ 

s/'d ' ’ 


(422) 


* Cambria Handbook, p. 29. 







i8o 


CIVIL ENGINEERING. 


in which s s " = allowable unit stress in shearing of the material; 

t/' = net depth of the web in inches; 
t = thickness of the web in inches; 

(F s ) w = vertical shear at the support or the maximum ver¬ 
tical shear in pounds. 

The net depth of the web is the gross or full depth assumed, 
less the sum of the diameters of the rivets in a vertical section. 
The sum of the diameters of the rivets in the stiffener must be 
subtracted from the gross depth to determine the net depth. 

The thickness of the web should be the larger of the values 
determined from the above formulas, and for ordinary girders 
should be from | to b inch. 

The Web-rivets. — The web-rivets are the rivets which unite 
the angles and web. Their purpose is to so unite the web and 
flanges that the total flange stress from section to section shall 
vary as closely as possible with the bending moment. 

Since the total flange stress can vary only at the web-rivets, the 
horizontal shear on any web-rivet must be equal to the difference 
of the flange stresses on either side of the rivet. This is equal 
to the difference between the bending moments at the given 
rivet and at the preceding rivet, divided by the effective depth of 
the girder. Hence 

qf' _ 

s s A’"*=(s l -s 2 )A' = --T...(423) 

Li 1 

in which s s = unit horizontal shear in any web-rivet; 

A n '= area of cross-section of a web-rivet in square 
inches; 

d\ = effective depth of the girder; 

$i = the unit flange stress on the side of the rivet 
towards the dangerous section; 
s 2 = unit flange stress on the opposite side; 

H' = net flange area; 

M f = bending moment at preceding rivet on the side 
of the dangerous section; 

M" = bending moment at the rivet. 

From the principle partially developed on page 90, that the 
bending moment at any section of a beam without weight is 

. . * Tf bearing value of rivet is less than shearing value, sbbt must be substi¬ 
tuted for s s A 




SOLID BUILT BEAMS , I BEAMS , AND PLATE GIRDERS. 181 


equal to the bending moment at any other section plus the mo¬ 
ment of the vertical shear at the second section with respect to 
the first, we have 


M' = M" + F s a, or M’ - M" = F s a, . . . (424) 


in which V s — vertical shear at rivet where the moment is M "; 

a = distance between this rivet and its adjacent one, or 
the pitch at this point of the girder. 

Substituting this value in equation (423) we have 


s s A'"di = V s a, or 


s,A'”di 



(425) 


In this equation if we make A'" constant, that is, make all 
the rivets of the same size, and make s s = s s ", the safe unit stress 
in shear, it is evident that a, the distance between rivets , must 
vary inversely with F s , the vertical shear. Where the vertical 
shear is greatest, as at the ends of a girder uniformly loaded and 
resting on end supports, the web-rivet spacing must therefore be 
least; and at the dangerous section, where the vertical shear is 
least, the rivet-spacing must be greatest. 

The depth between the upper and lower web-rivets is fre¬ 
quently substituted in this formula for the effective depth; this 
increases slightly the factor of safety. 

As it is not economical in practice to vary the rivet-spacing 
at every point, it is customary to divide the girder into panels 
and make the spacing the same throughout each panel, but to 
make the spacing in the different panels unequal. The maxi¬ 
mum rivet-spacing is also governed by the rule that the maximum 
pitch in a girder must nowhere exceed six inches. 

If a uniform load is placed on the upper flange, it may be trans¬ 
mitted to the web through the web-rivets. In that case the shear¬ 
ing and bearing resistances of these rivets in each panel should be 
sufficient to resist the resultant of the load on the panel and the 
horizontal shear in its web-rivets. 

The Flange-rivets. — The flange-rivets are those connecting 
the flange-plates and angles. As the flange-plate stresses must 
also vary with the bending moment, the same conditions govern 



182 


CIVIL ENGINEERING. 


the horizontal stresses in both the flange- and web-rivets. The 
flange-plates, however, resist only a part of the total flange stress, 
hence the stress on the flange-rivets is less than the stress on the 
web-rivets; the flange-rivets are, however, in single shear, while 
those in the web are in double shear. The spacing of the flange- 
rivets is also governed by the condition that the maximum spac¬ 
ing shall not exceed six inches. 

The flange-plates make the effective depth of the girder greater 
than it would be were they omitted. They also reduce the unit 
flange stress by distributing the stress over a larger area of cross- 
section. By varying the thickness of the flange-plates, making 
the thickness greatest where the bending moment is greatest, and 
the thickness least where the bending moment is least, the 
girder may be made approximately a beam of uniform strength. 
Each flange-plate is made longer at each end by twice the rivet¬ 
spacing than actually required by theory. 

Stiffeners. — The web-stiffeners shown in Fig. 40 are small 
angles or tees employed in pairs on opposite sides of the web, and 
are riveted to the web and to each other. The stiffeners serve to 
transmit concentrated loads from the upper flange to the web, and 
also to prevent the web from buckling. For the former purpose 
they must be placed at the supports and wherever the load is 
concentrated; for the latter, they are placed at intervals equal to 
the depth of the girder, but not exceeding five feet. 

To determine whether stiffeners are necessary to prevent the 
buckling of the web, the value of is deduced from the following 
formula; if it exceeds the unit shearing stress in the web previously 
determined, it is assumed that the web requires stiffeners to prevent 
its buckling: 



1 “.000 


1 + 


d " 2 ’ 
3000/ 2 


(426) 


in which d" = depth of web, and t = thickness of web. 

Each pair of stiffeners is made strong enough to support the 
entire vertical shear at its section without exceeding the value of 
in the following formula: 




SOLID BUILT BEAMS, I BEAMS, AND PLATE GIRDERS. 183 


in which / = length of stiffener in inches; 

r = its least radius of gyration in inches. 

Lateral Stiffness.—The compression-flange of a girder resting 
on end supports is a long column supported along one side. It is 
evident that it requires lateral support to prevent it from buckling. 
This support is secured by connecting the compression-flanges of 
consecutive girders bv suitable braces, whenever the length of 
the compression-flange exceeds sixteen times its width. 

Third Method.—In the third method the flange area is not so 
great as in the second, since the web is assumed to bear part of 
the longitudinal stresses. 

Having determined the area of the web as before, the area of 
each flange is deduced from the formula 

.(428) 

in which s" = safe unit stress; 

di = effective depth of girder in inches; 

A'= area of each flange in square inches; 

H" = area of web in square inches. 

This formula is derived from the general formula 



1st. By assuming that 2,4V 2 , the moment of inertia of the 

A'di 2 

two flanges, is equal to ——, or that the effective depth oj the 

girder is equal to twice the radius oj gyration oj the jlanges about 
the neutral axis. 


A"d " 2 

2d. By assuming that ———, the moment of inertia of the 


12 


A"dv 2 


web about the neutral axis, is equal to ■ -, or that the effec¬ 

tive depth oj the girder is equal to the gross depth oj the web , d". 

3d. By assuming that di = 2y', or that the effective depth oj 
the girder is equal to its gross depth. 

Each of these assumptions introduces a slight error, since 
di<2r, di<d" unless the flange-plates are exceptionally large, 

and — < y . 

2 







184 


CIVIL ENGINEERING. 


Under these assumptions we have 


I_ 

y'~ 



2 


Since the web resists some of the longitudinal stress, the web- 
rivets are subjected to less horizontal shear than in the second 
method, and the rivet-spacing may be increased in the ratio of 
the total longitudinal stress to the longitudinal stress borne by 
the flanges. 

The strength of the web and the web-splicing must be greater 
than in the second method in order to carry the horizontal stress 
in addition to the vertical shear. 

Box Girders.—Box girders are formed by placing two or 
more rolled I beams or plate girders side by side and connect¬ 
ing them by flange-plates which extend over and under all the 
beams or girders. They are employed when great strength is 
required. 

Application.—Let it be required to design, with the aid of a 
structural-metal handbook, a steel plate girder, 30 feet long be¬ 
tween supports, which is to carry a load of 2000 pounds per linear 
foot, including the weight of the girder. 

Specifications.—1° Web and Stiffeners .—Depth of the web 
equals T y of the span. Allowable unit stress in simple shear 
equals 12,000 pounds. The vertical shearing stress is to be 
borne entirely by the web-plate. The web must be stiffened at 
intervals not exceeding the depth of the girder, or a maximum 
of five feet wherever the unit shearing stress is greater 
than 12,500 — 90 H,* in which H equals the ratio of the depth of 
the web to the thickness. The edges of the web-plate must not 
project beyond the faces of the flange-angles, nor shall they be 
more than J inch inside these faces at any point. 

The web-plate must have stiffeners over bearing-points and 
at points of local concentrated loads. Such stiffeners must be 
fitted at their ends to the flange-angles over the bearing-points. 


* A right-line form of equation (426). 








SOLID BUILT BEAMS , / BEAMS , PLATE GIRDERS. 185 


The stiffeners are to be made of angles with equal legs. All 
stiffeners must be capable of carrying the maximum vertical shear 

without exceeding the unit stress in the formula s =12,000 ——, 

^ C ' y ' 

in which / = length of stiffener in inches, and r = its radius of 
gyration in inches about an axis parallel to the web of one angle. 
Each stiffener must be fastened to the web by enough rivets to 
transfer the total vertical shear at that point to or from the web. 

2 0 Rivets. — Allowable unit stress in single shear = 6000 
pounds. Unit bearing value for rivets = 12,000 pounds. The 
unsupported width or distance between rivets of flange-plates 
in compression shall not exceed thirty times the thickness of 
the plates. The rivets used shall be f inch in diameter. The 
diameter of rivet-holes to be deducted in order to obtain net 
sections shall be assumed as -J inch. 

3 0 Flanges. —Allowable unit flange stress = 13,000 pounds. 
The flanges shall be proportioned under the supposition that 
they carry the entire bending or longitudinal stress. The com¬ 
pression-flange shall be made of the same gross area as the tension- 
flange. All the joints in riveted flanges shall be fully and sym¬ 
metrically spliced. At least one-half of the flange section shall 
be angles. Flange-plates must extend beyond their theoretical 
limits two rows of rivets at each end. The flange-plates must 
be limited in width so as not to extend beyond the outer lines 
of rivets connecting them with the angles more than five inches 
or more than eight times the thickness of the plate. The com¬ 
pression-flange must be stayed against transverse crippling when¬ 
ever its length exceeds sixteen times its breadth. 

Maximum Stresses.—The dangerous section will be at the 
middle point, where the maximum bending moment is 

Wl 12 r , j \ 

M m = — = 2000X30X30X^ = 2,700,000 (inch-pounds). 

30 X 2000 

The maximum shear equals '-- — 30,000 pounds. 

£ 

Tension-flange.— Assume effective depth of girder = d x =23.5 
inches. Then 




j86 


CIVIL ENGINEERING. 


A's"d\=A' X 13,000 X 23.5 = 2,700,000; 


hence 


2,700,000 

13,000X23.5 


8.87 square inches. 


Try two angles 5"X3"X|" *. 5 - 7 2 sq. inches gross area 

one flange-plate 12" X f". 4 - 5 ° “ “ “ 

Total. 10.22 

Deduct J" rivet-holes con¬ 
necting flange-plate and 
angles. 1.32 

Net area. 8.90 sq. inches, as required 

To Find the True Value of the Effective Depth.—The moment 
of one flange about the neutral axis must be equal to the moment 
of its angles plus the moment of its flange-plate. 


C. of gr. of flange-plate 
from neutral axis 
C. of gravity of angles 
from neutral axis 
C. of gravity of rivet- 
holes from neutral 
axis 


J =12 in.-f J- in.-ftV in. = 12.31 inches; 

■* 

• =12 in.+ J in.—0.70 in.f = 11.43 inches; 
► = 12 in. +J- in. = 12.12 inches. 


12 inches = one half the depth of the web; J inch = distance 
from edge of web to flange-plate; and 0.70 inch = the distance 
from center of gravity of angles to their backs. Hence 


d 1 

8.90 X— =4.50X12.31 + 5.72X11.43 — 1.32X12.12. 

Solving we find d\ = 23.54 inches. This is greater than the 
assumed value, 23.5, and is therefore on the side of safety. 


* Cambria Handbook, p. 174, A 101. 
t Ibid., p. 174, column 6. 













SOLID BUILT BEAMS, I BEAMS, AND PLATE GIRDERS. 187 


Compression-flange.—This will be made of the same gross 
area as the tension-flange. 

Design of Web.—1° For Simple Shear .—The web is 24 inches 
deep, and a thickness of f inch will be assumed, as the girder 
is a light one, and this is a thickness commonly used. The 
stiffeners at the ends must carry a compressive force of 30,000 
pounds from the points of support to the web. This will require 
nine j-inch rivets, since the bearing value of a single rivet in a 
-f-inch plate is 3375 pounds.* The total number of rivet-holes 
to be deducted from the gross area of the web is therefore nine, 
in which are included the two rivets required to connect the flanges 
with the web at the same cross-section. The net section of the 
web will then be (24 — 9x1)1 = 6.05 square inches, and is amply 
strong to carry the maximum shear of 30,000 pounds, since 
6.05 X 12,000 is greater than 30,000. The seven rivets connect¬ 
ing the stiffeners with the web directly will have a pitch of 
24-OX3) 


7 


= 2.6 inches, which lies between the allowable limits 


of three diameters and six inches, and may be adopted. 

Each angle forming the stiffener carries a load of 15,000 

pounds. Try two angles 2JX2JXA inches,! for which r=. 76. 

55 l T , • , i.S,ooo 

= 10,300. ihe area required = 


Hence 5 =12,000 

c r " 10,300 

1.46; the angle selected has an area of 1.47 square inches and 
will be used. As the combined thickness of the angles is greater 
than the thickness of the web, the bearing area will be sufficient. 
2 0 To Resist Buckling .—The allowable unit stress in the 

90X24 


web is 12,500 — 90^=12,500 — 


3 

8 


= 6740 pounds per square 


50,000 

inch. The actual stress is -7 - , which is less than 6740, and 

6.05 

hence no stiffeners are required except at the ends. 

3 0 Test jor Bearing .—The bearing value has already been 
found satisfactory for the rivets connecting the end stiffeners 
with the web, and the calculation for the number of rivets con¬ 
necting the flange-angles with the web will indicate whether 
it is sufficient for them. 

* Cambria Handbook, p. 308, first table, 
f Ibid., p. 166, A 17. 









188 


CIVIL ENGINEERING. 


Rivets Connecting Flange-angles and Web. — The bearing 
value of a j-inch rivet in a f-inch plate is 3375 pounds, and is 
less than its value in double shear; the bearing value will there¬ 
fore determine the rivet-spacing. 

Since the maximum bending moment is 2,700,000 (inch-lbs.) 
and the effective depth is 23.5 inches, the maximum flange stress 

is — 0Q, ° 0 ° , or 115,000 pounds approximately. This is equal 
—3 * 5 

to the horizontal shear on all the web-rivets of each flange in 

each half of the girder. The total vertical shear on the web- 

rivets of each flange in one half the girder is 30,000 pounds. 

The resultant shear is V(i 15,ooo) 2 + (30,ooo) 2 = 118,000. The 

total number of web-rivets required in each half of each flange 

. ^ r 118,000 

is therefore -= 35. 

3375 

To determine the rivet-spacing we have 


a = 


3375^1 

Vs 


= 2.6 inches, 


in which d 1 = 23.5; 

V s = 30,000; 

a = rivet-spacing at the supports. 

To secure rivet-spacing in even inches, let the spacing be 2J, 
3, 4, 5, and 6 inches. If in the formula above we substitute for 
a, 3 inches, and solve with respect to V s , we shall have 7 S = 26,438 
pounds approximately. Hence from the point of the girder 
where the vertical shear is 26,438 pounds the rivet-spacing towards 
the middle point need only be 3 inches. Since the load is 2000 
pounds per lineal foot, this is about 2 feet from the end of the 
girder. 

Similarly the rivet-spacing need not exceed 4 inches from 
the point where 73 = 19,829 pounds. This is about 5 feet from 
the end. For 5-inch pitch 73 = 15,683 pounds, and for 6-inch 
pitch 73 = 13,219. 

Beginning at one end, the rivet-spacing will be 2\ inches for 
a length of 2 feet; then 3 inches for a length of 3 feet additional, 
etc.; over the middle 13 feet of the girder the minimum allowable 
spacing for girders, or 6 inches, will be employed. In each half 







SOLID BUILT BEAMS, l BEAMS, AND PLATE GIRDERS. 189 

of the girder there will be 10+ 12 + 6 + 6 +8^ = 42^ web-rivets in 
each flange, which is more than required. 

Rivets Connecting Flange-plates and Flange-angles. —These 
rivets are in single shear, but carry less stress than those con¬ 
necting the flange-angles with the web. If, then, they are given 
the same pitch as the web-rivets, there being twc flange-rivets 
for each web-rivet, they will safely transmit the stress. 

Flange-plates. —These may be omitted between the supports 
and the section where 5" multiplied by the area of two angles 
multiplied by d\=M. In this expression d\ is the distance be¬ 
tween the centers of gravity of the flanges when the flange-plates 
are omitted, and equals 24 + \ — 2 X.70 = 22.85. 

Substituting numerical values in s"A'di =M we have 

2000X 2 

13, 000 x 5-72 x 22.85=3°, oo° x- 7T^r’ 

whence x = f jo and 290, and the theoretical length of the flange- 
plates is 220 inches. This length must be increased so as to 
have two additional rows of rivets at each end, and so adjusted 
in length that the rivets connecting the plates with the angles 
will break joints with those connecting the angles with the web. 
This makes the length to the nearest full inch 240 inches or 
20 feet. 

Summary. —The girder will then be made up as follows: 


Four angles 5" X 3" X f". 1176.0 pounds * * * § 

One web X24"X30'. 918.0 “ f 

Two flange-plates X 12"X20'. 612.0 “ f 

Four angles 24" X 2J" Xte" . 4 °-° “ t 

Heads of 375 rivets. 124.0 § 


Total. 2870.0 pounds 


The weight of the girder is therefore about 5 per cent of the 
applied load. 


* Cambria Handbook, p. 174, fourth column, 

t Ibid., p. 399, last column. 

% Ibid., p. 166, A 17, fourth column. 

§ Ibid., p. 322, bottom of page. 













190 


CIVIL ENGINEERING. 


Girder Considered as a Solid Beam.—To find the safe load 
under this supposition the moment of inertia of the cross-section 
about the neutral axis of the section may be calculated as fol- 

MY 

lows and substituted in the formula s" = 13,000 = —in which 


y 


, 24+j + f 


=I2J". 


The moment of inertia of the angles and 


flange-plates is derived from the formula Ar 2 = Ad 2 + Ar' 2 . 


Ar ' 2 = i of four angles about axes through center 
of gravity parallel to neutral axis of 

girder. 

A / 2 = i of two flange-plates about axes through 
center of gravity parallel to neutral axes 

of girder. 

Ar ' 2 = I of web about neutral axis of girder.... 
d V 2 

A --= 11.44X (n.42) 2 . 

4 

dY ' 2 

A—=gX(i 2 .T ) i ) 2 . 

4 

I of four holes to be deducted = 2§X (12J) 2 . 


8.16* 

neglected 
432.00 t 

1493.00 

1364.0 

386.0 


Total about. 2911.00 


Therefore 


My 

13,000= ~y 


Wly 4; Wy 45 W X12.5 


SI 


2911.00 


whence = 67,277 app. Therefore the girder would be safe con- 


7277 

sidered as a solid beam under a load -- 

00,000 

than that for which it has been calculated. 


= 12 per cent greater 


* Cambria Handbook, p. 174, A 101, seventh column, 
f Ibid., p. 184. 


















CHAPTER X. 


DETERMINATION OF STRESSES BY ANALYTIC METH¬ 
ODS IN SIMPLE TRUSSES RESTING ON END SUP¬ 
PORTS AND CARRYING UNIFORM DEAD LOADS. 

A truss is a jrame designed , like the simple beam and plate 
girder , to support weights and to transfer their effects to lateral 
supporting walls. A truss is therefore subjected to the bending 
moments and to the vertical and horizontal shears of the forces 
acting on it. (Fig. 41, p. 194.) 

In the simple beam each fiber is subjected to a longitudi¬ 
nal and a shearing stress; in a girder the fibers of the flanges 
are assumed to be subjected to longitudinal stresses only, and 
the fibers of the web to shearing stresses only; in a truss all 
the fibers are subjected to longitudinal stresses only. 

Trusses are used principally to support the floors and floor 
loads of bridges, and the floors, floor loads, and roofs of build¬ 
ings when the supporting walls are widely separated. Floor- 
trusses are subjected to weights or vertical forces which may 
be either stationary or moving; the former are called dead and 
the latter live loads. Roof-trusses are subjected to weights or 
vertical forces, and to wind pressures or non-vertical forces. 

Trusses may be divided into two general classes, those whose 
upper and lower pieces are parallel and horizontal, as the floor- 

trusses shown in Figs. 41, 4 3 > and 45 > and those in which the 
upper and lower pieces are not parallel, as in the roof-trusses 
shown in Figs. 44 and 46. The separate pieces of a truss are 
called the members; the upper and lower members are the chord 
members , and the intermediate ones the web members or braces. 
In a roof-truss the upper chord is also called the principal 1 after. 
Web members subjected to tensile stress are called ties, those 
subjected to compressive stress, struts. 

Trusses with Parallel Horizontal Chords— A plate girder 

may be transformed into a truss with parallel horizontal chords 

191 


192 


CIVIL ENGINEERING. 


by simply substituting for its web a system of web members 
which are fastened to each other and to the flanges by pin or 
riveted joints and divide the web area into a number of triangles, 
Fig. 41. The joints between the chords and web members are 
called the panel points. 

In making the transformation, the flanges of the girder 
become the chords of the truss, which, like the flanges, are 
designed to resist the entire bending moment of the forces act¬ 
ing on the truss. The stresses in both flanges and chords are 
similar longitudinal stresses; the only difference between them 
being that the flange stress changes at every web-rivet and is 
uniform between rivets, while the chord stress changes at every 
panel point, and is uniform between panel points. The web 
members perform in the truss the same function as the web in 
the girder; they are designed to resist the entire vertical and 
horizontal shear of the forces which act on the truss. The hori¬ 
zontal component in each web member must therefore be equal 
to the horizontal shear, and the vertical component to the verti¬ 
cal shear, at the section of the truss. In the girder the loads 
and horizontal shears are transmitted from the rivets of the 
upper flange to the web-plate and produce shearing stresses in 
it; in the truss the loads and horizontal shears are transmitted 
from the panel points of the chords to the web members, and 
produce longitudinal stresses in them. The pins or rivets of 
the truss perform the same function as the web-rivets in the 
girder; the web-rivets connect the chord and web members and, 
bv transferring the excess longitudinal chord stress to the web 
members, allow the cross-sections of the' different members of 
the chords to vary with the bending moment. When the panel 
points in a chord are uniformly spaced the distance between 
them is called a panel length. 

The simplest forms of trusses with parallel chords are the 
Warren triangular truss, Fig. 41, the Pratt panel truss, Fig. 43, 
and the Howe panel truss, Fig. 45. In the Pratt truss the verti¬ 
cal web members are struts, and in the Howe, ties. In the fig¬ 
ures the tensile members are shown with light lines, and the 
compression members with heavy lines. 

Trusses without Parallel Horizontal Chords.—Whenever the 
chon^l of a truss is inclined to the horizontal it must transmit 


DETERMINATION OF STRESSES BY ANALYTIC METHODS. 193 

a portion of the vertical shear and hence relieve the web members. 
The chords of such a truss are not therefore designed to resist 
the bending moment only, nor are the web members designed 
to resist the entire shear. 

Methods of Determining the Stresses in the Members of a 
Truss.—As in the case of beams and girders, it is assumed that 
the loads or forces which are transmitted by a truss are coplanar 
forces which act in the plane containing the axes of the members. 
It is also assumed that the forces act only at the panel points 
of one or both of the chords. 

There are two general methods of determining the stresses 
in the members of a truss, the analytic and the graphic. The 
analytic may be subdivided into the analytic method oj concurrent 
jorces and the analytic method oj moments. 

Analytic Method of Determining Reactions.—In all trusses 
resting on end supports the reactions at the supports are assumed 
to act along lines parallel to that of the resultant of the other 
external forces, unless the action lines of the reactions are fixed 
by special conditions. If the truss rests on a roller at either end, 
to allow for the expansion and contraction of its members due to 
changes of temperature, the reaction at that end is assumed to 
be normal to the surface on which the roller rests. 

The two reactions with the resultant oj the other jorces will 
therejore always jorm a system either oj parallel or concurrent 
jorces in equilibrium. Having constructed this resultant, the 
intensities and directions of the reactions may always be deter¬ 
mined from the general conditions of equilibrium of a system 
of coplanar forces, by taking moments with respect to the points 
of support. 

Analytic Method of Concurrent Forces.—This method depends 
upon the following principles: 

If the truss is in a state of rest, the external forces acting on 
the truss as a whole, and the forces and stresses acting at each 
panel point , will form separate systems of forces in equilibrium. 
The system of forces at each panel point being a system of con¬ 
current forces, the only requisites for equilibrium are that the 
algebraic sums of the horizontal and vertical components shall 
each be equal to zero. 

In any system of concurrent forces in equilibrium in which 


194 


CIVIL ENGINEERING. 


the lines of direction of all the forces are given, the intensities 
of two and of not more than two unknown forces may be deter¬ 
mined if all the others are given. This is evident since, if we 
find the resultant of the known forces and its components in the 
directions of the unknown forces, equilibrium is possible only 
when the unknown forces are equal in intensity and opposed 
in direction to these components. 

In any system of concurrent stresses in a truss the stresses 
acting towards the common point of application are compressive , 
and those acting away from the point are tensile; and, conversely, 
a compressive stress in a piece must be assumed as acting towards 
the common point of application, and a tensile stress as acting 
away from it. 

J 

Stresses in Trusses with Parallel Chords.—Let these prin¬ 
ciples be applied to the problem of determining the stresses in 
the Warren truss shown in Fig. 41 due to a load of W at each 
panel point of the lower chord. 



Fig. 41. 


If it is assumed that there are stresses in all the members of 
this truss, it is seen that the only panel points where there are 
but two unknown stresses are at the points A and A hence 
we must begin the solution at one of these points. 

Panel Point A.—At the panel point A if we resolve each 
force and each stress into horizontal and vertical components, 
and indicate the forces acting upwards and those acting to the 
right as positive, and those acting downwards and those acting 
towards the left as negative, we shall have the following table, 
in which the given stresses are in Roman and the required 
stresses in Italic characters. 

In the first line we write the known force and its horizontal 
and vertical components. In the second line we write the 
unknown force whose vertical component must be equal to the 






DETERMINATION OF STRESSES BY ANALYTIC METHODS. 195 


Force. 

Vertical 

Components. 

Horizontal 

Component. 

Total Intensity. 

Character. 

2 \V 

Stress BA 

“ AC 

+ 2W 

-2\V 

0 

O 

A-2W tan (j) 

— 2W tan $ 

2 \V 

2}V 

COS <p 

2W tan <p 

Reaction 

Compressive 

T ensile 


vertical component of the known force. At the point A this 
must be the stress BA , since the stress CA, if resolved into hori¬ 
zontal and vertical components, will have a vertical component 
of zero. The vertical component must act downwards, hence 
the stress is compressive and its horizontal component acts 
towards the right or is positive. Knowing its vertical com¬ 
ponent and the angle <p, the stress BA is fully known in intensity 
and direction, and we can write out its horizontal component, its 
total intensity, and its character. In the third line we write the 
force whose intensity must be equal and contrary to the hori¬ 
zontal component of the now known forces. Since the hori¬ 
zontal component of AC is negative, it acts towards the left and 
away from A ; the stress is therefore tensile. If the table is 
correct, the sum of the vertical and horizontal components of 
the forces or stresses will be separately equal to zero. 

Panel Point B.—At the panel point B the compressive force 
AB is fully known and acts towards the point B or upwards 
and to the left. Its vertical component must therefore be 
positive and its horizontal component negative. Hence we may 
write out the following table: 


Force. 

Vertical 

Component. 

Horizontal 

Component. 

Total Intensity. 

Character. 

Stress AB 

+ 2\V 

— 2W tan cp 

2 W 
cos <p 

Compressive 

Stress BC 

“ DB 

-2lV 

0 

— 2IV tan <p 
p\V tan p 

2\V 

COS cj) 

4W tan <p 

T ensile 

Compressive 


As the vertical component of BC must be negative to counter¬ 
act the vertical component of AB, the stress BC acts downward 
and away from B, hence its horizontal component acts towards 
































196 


Cl NIL ENGINEERING 


the left and is negative, and the stress itself is tensile. The 
intensity of the stress in DB must be equal to the sum of the hori¬ 
zontal components of the stresses AB and BC, and act in a con¬ 
trary direction or towards B. It is therefore a compressive 
stress. 



Fig. 41. 


Panel Point C.—At this panel point we know both the tensile 
stresses BC and CA and the force IF; hence they are placed in 
the first three lines of the table: 


Force. 

Vertical 

Component. 

Horizontal 

Component. 

Total Intensity. 

Character. 

Stress CB 

+ 2 W 

+ 2W tan <j> 

2 W 

Tensile 

Stress CA 

O 

+ 2W tan p 

cos 9 

2W tan (b 

< ( 

W 

-w 

0 

W 

Applied load 

“ DC 

— IF 

+ W tan (ji 

IF 

COS 4> 

Compressive 

“ CE 

0 

—5IF tan <p 

5IF tan <p 

T ensile 


The vertical component of DC must be equal to —IF, hence 
the force acts towards C, is compressive, and has a positive hori¬ 
zontal component. The stress CE must be negative; it there¬ 
fore acts away from the point C and is a tensile stress. 

Panel Point D.—At this panel point we know the compres¬ 
sive stresses BD and CD. 


Force. 

Vertical 

Component. 

Horizontal 

Component. 

Total Intensity. 

Character. 

Stress BD 

O 

— 4W tan (p 

4W tan cf> 

Conijiressive 

“ CD 

+w 

— W tan (j> 

W 

COS </) 

IF 

( ( " 

Stress DE 

— W 

— TF tan <p 

COS (f> 

T ensile 

“ FD 

0 

-f 6 W tan (ji 

6 IF tan p 

Compressive 











































DETERMINATION OF STRESSES BY ANALYTIC METHODS. 197 

The vertical component of DE must be equal to — IF, hence 
the stress acts away from the point D, is tensile, and has a nega¬ 
tive horizontal component. The stress FD must act towards 
D and is compressive. 

Panel Point E.—At this point we know the tensile stresses 
ED and EC and the force IF. 


Force. 

Vertical 

Component. 

Horizontal 

Component. 

Total Intensity. 

Character. 

Stress ED 

+w 

-f- W tan <£ 

w 

Tensile 

“ EC 

0 

+ 5W tan <f) 

cos 9 

5W tan 

( ( 

W 

-W 

0 

W 

Applied load 

Stress EE' 

0 

0 

0 


“ EE' 

0 

— 6\V tan (j) 

6 IV tan ([> 

T ensile 


The only forces acting at the point E which have vertical 
components are the stresses ED and EF and the force W. Since 
the vertical component of the stress DE is exactly equal and 
opposed to the force IF, the vertical component of the stress in 
EF must be equal to zero. If its vertical component is zero, the 
stress itself must be equal to zero, since its action line is not 
horizontal. The stress in EE' must be negative and equal to 
the stress in EC and the horizontal component of ED; the stress 
in EE' must therefore be tensile. 

As the truss is symmetrical, the stresses in the members of 
the left half must be equal to the stresses in the corresponding 
members of the right half. 

The stresses may then be tabulated as shown on next page. 

If we examine the table it will be seen: 

I. That the stresses in the web members are alternately 
compressive and tensile, and that between the points of appli¬ 
cation of the external forces the web members are subjected 
to the same amount of stress. This results from the fact that 
there is no change in the vertical or horizontal shear between these 
points. 

II. That the stresses in the upper chord are all compres¬ 
sive and those in the lower chord all tensile, and that these stresses 
increase as we approach the middle of the span. This was to 
be expected, as the longitudinal fiber stresses in a beam or the 














198 


CIVIL ENGINEERING. 


Stresses. 

Compressive. 

Tensile. 


Upper Chord. 

• 

Stress BD 

4W tan 


“ FD 

6W tan (f> 



Lower Chord. 


Stress AC 


2W tan <j> 

“ CE 


5W tan (j) 

“ EE' 


6W tan <j> 


Web Members. 


Stress AB 

2\V 


“ BC 

COS 0 

2W 


W 

COS <}) 

“ DC 




cos 9 




W 

“ DE 


cos 

“ EF 

0 

0 


fiber stresses in the flanges of a girder resting on two points of 
support are compressive above the neutral axis and tensile 
below, and the bending moment is a maximum at the middle 
point of a uniformly loaded beam resting on two end supports. 

If the numerical coefficients of the forces and stresses are 
written on the diagram of the truss, we shall have the diagram 
shown in Fig. 42. 



Fig. 42. 


If this diagram is carefully examined, it will be seen that 
at every panel point the sum of the numerical coefficients of the 
stresses which tend to move the point upwards are exactlv equal 
to the sum of the coefficients of the stresses which tend to move 
it downwards; and that the sum of the numerical coefficients 
of the stresses which tend to move it to the left is exactly equal 
to the sum of the coefficients of the stresses which tend to move 
it to the risrht. 






























DETERMINATION OF STRESSES BY ANALYTIC METHODS. 199 


This enables us to write out the numerical coefficients of 
the stresses of any truss with parallel chords and web members 
making the same angles with each other, provided we are care¬ 
ful to indicate by arrow-heads the direction of the stress in 
each piece as soon as determined. The arrow-heads will also 
indicate the character of the stress. A compressive stress must 
be indicated as acting towards both extremities of the member 
in which it is found, and a tensile stress as acting away from both' 
extremities of the member in which it is found. 

In the Pratt truss, Fig. 43, at the panel point A the reaction 






A 

r< >/ 

A ' 

r<- >> 

y 

X r v 

> 

5 

CM 

^ £ 

V 2\ 

' 

N 2\ 

t G 8 v 

N 2\ 

,i 5 

✓ 2\ 

\c { 

N 5V\ 


Pratt Truss. 

Fig. 43. 

5 IF tends to move the point A upwards; it is restrained only 
by the member AB ; the stress in AB must therefore be com¬ 
pressive and its numerical coefficient must be 5. 

The stress in AC must be zero, since there is no opposing 
force in a horizontal direction at A. 

At the panel point B the known compressive stress in A B 
tends to move the point B upwards, but it is restrained by the 
piece BC\ hence the numerical coefficient of the stress in BC 
is equal to that of AB, or is 5. The stress in BC must act down¬ 
wards and be tensile. The stress tending to move B in a hori¬ 
zontal direction to the left is BC, but it is restrained by the 
stress BD; hence the stress in BD must be compressive and its 
numerical coefficient is 5. 

At the panel point C the known stress tending to move the 
point C upwards is the tensile stress BC with a numerical coeffi¬ 
cient of 5; the known force tending to move it downwards is the 
applied load 2IF; hence the stress in DC must act downwards or 
towards C, and be compressive and have a numerical coefficient 
of 3. The known stresses tending to move the point C to the 
right are the stresses CB and CA; it is restrained by the 
stress in CE, which must act towards the left. The stress 
in CA is zero; hence the stress in CE acts towards the left, 











200 


CIVIL ENGINEERING . 


is tensile, and has a numerical coefficient equal to that of CB, 
or 5. 

In the same manner we may determine the numerical coeffi¬ 
cients of all the other stresses. 

To determine the actual stresses the 

Coefficients in the chord members should be multiplied by 

W tan <j). 

“ “ “ diagonal web members should be multi- 

IT 

plied by- 7. 

r cos <j> 

“ “ “ vertical web members should be multi¬ 

plied by W. 

In these expressions cf) is the angle CBA. 

In writing out the stresses in a truss by this method, it is 
usually simpler to write out first the stresses in the web mem¬ 
bers, since the numerical coefficients of these stresses are the 
same as the numerical coefficients of the vertical shear. Having 
the stresses in the web members, those in the chords are easily 
determined. Thus in the Pratt truss, Fig. 43, the vertical shear 
between A and C is 5IT, hence the numerical coefficient of the 
stresses in AB and BC is also 5; the vertical shear between 
C and E is 5IT — 2lT = 3lT, hence the numerical coefficient of 
the stresses in CD and DE is also 3; the vertical shear between 
E and G is 5IT — 2IT — 2lT= IT, hence the numerical coeffi¬ 
cient of the stresses in EF and FG is also unity. 

It will be observed that if in Figs. 41 or 42 we cut the truss 
by a plane which intersects one diagonal web member only, 
the sum of the numerical coefficients of the compressive stresses 
in the members cut will be equal to the sum of numerical coeffi¬ 
cients of the tensile stresses. Thus if, in Fig. 43, we pass a verti¬ 
cal plane between E and C, cutting the members EC, ED, and 
DF, we have the sum of the numerical coefficients of the tensile 
stresses in CE and DE equal to the numerical coefficient of the 
compressive stress in FD. This affords a means of checking 
results obtained when the stresses are written out as described 
above. 

Stresses in Trusses without Parallel Chords.—The solution 

by analysis of concurrent forces may also be applied to trusses 
without parallel chords. 



DETERMINATION OF STRESSES BY ANALYTIC METHODS. 


201 



p 

2W 

JVw 2W 

B J<: 


// / 


- > < c 

, >■ " , 


Ki.ig-post Truss. 
Fig. 44. 


Let it be required to determine the stresses in the truss 
shown in Fig. 44 by the analysis 2 w 

of forces. 

It will be observed that we may 
determine the stresses in AB and 
AC in the same manner in which 

we determined the stresses in the 3W 3w 

corresponding pieces of the truss 
with parallel chords, by resolving 
the forces acting at the point A into horizontal and vertical compo¬ 
nents. However when we reach the panel point B we shall have 
both horizontal and vertical components in both of the unknown 
members CB and DB. The resolution of the stresses into hori¬ 
zontal and vertical components will not aid us in the solution 
of the unknown stresses at that point. The same method of 
determining the stresses may, however, be utilized if we take 
our components at A, C, and D vertical and horizontal, and 
at B normal and parallel to the chord AD. The angle between 
the chords is 90 0 —</>. 

As before, the stresses acting upwards or to the right will 
be considered positive, those acting downwards or to the left 
will be considered negative. 


Panel Point A. 


Stress. 

Vertical 

Component. 

Horizontal 

Component. 

Total Stress. 

Character. 

3 W 

+ 3 W 

O 

3 W 

Reaction 

AB 

- 3 W 

+ 3IV tan (f) 

3W 

cos d> 

Compressive 

AC 

0 

— jlV tan <p 

jW tan cp 

T ensile 


Panel Point B. 


Stress. 

Normal 

Component. 

Parallel Component. 

Total Stress. 

Character. 

2 W 

— 2W sin (p 

4 - 2W COS (p 

2 W 

Applied load 

AB 

0 

3 W 
cos (p 

3 W 

COS (p 

Compressive 

BC 

+ 2IV sin <p 

4- \y si” 2 <p — W cos 2 <p 

W 

Compressive 

cos p 

cos <p 

BD 

0 

COS (p 

2 \V 
cos <p 

f ( 










































202 


CIVIL ENGINEERING. 


/ 


The normal component of the stress in BC must be equal 
in intensity and opposite in direction to the normal component of 
2 W, which is 2 IT sin 0 . The parallel component of the stress in 
BC must therefore be equal to 2W sin 0 cot (180 0 —20), which 

cot 2 0 — 1 


is equal to — 2IT sin 0 cot 20. Cot 20 = 


stituting for cot 0 its value 


COS 0 
sin 0 


2 cot 0 


Sub- 


we have 


cos 2 0 — sin 2 0 


, sin 2 0 

cot 20=-- - 

2 cos 9 


sin 0 


cos 2 0 — sin 2 0 
2 cos 0 sin 0' 


Substituting this value of cot 20 in the term — 2 IT sin 0 cot 20, 

.... . W sin 2 0 — W cos 2 0 

the parallel component becomes- 


The stress in BC is equal to 


cos 0 
2W sin 0 


2 IT sin 0 


sin (180 0 —20) ~ sin 20 
Sin 20 = 2 sin 0 cos 0 , which substituted above makes the stress 

IT 

in BC equal to- r. 

^ cos 0 


Panel Point C. 


Stress. 

Vertical 

Component. 

Horizontal 

Component. 

Total Stress. 

Character. 


-w 

— W tan $ 

w 


BC 

COS <f) 

W 

Compressive 


+ W tan <f) 

B'C 

-w 

< ( 



COS (f> 


AC 

0 

+ 3W tan 0 

3W tan 0 

Tensile 

A'C 

0 

— j\V tan $ 

jlV tan $ 

Tensile 

DC 

+ 2 W 

0 

2\V 

( i 



Panel Point 

D. 


Stress. 

Vertical 

Component. 

Horizontal 

Component. 

Total Stress. 

Character. 

2W 

-2W 

O 

2W 

Applied load 

CD 

-2W 

O 

2W 

Tensile 

BD 

+ 2\V 

— 2W tan <f> 

2\V 

COS (j) 

Compressive 

B'D 

+ 2\V 

-\- 2 lV tan <f> 

2 \v 

COS 0 

Compressive 










































DETERMINATION OF STRESSES BY ANALYTIC METHODS. 203 
The following table gives the stress in each member: 


Piece. 

( 

Compressive. 

Tensile. 

AB 

BD 

BC 

AC 

DC 

3 W 
cos p 

2W 
cos p 

W 

COS p 

3W tan p 

2W 


The Analysis of Moments.—The solution by analysis of 
moments depends upon the principle that, since every member , as 
well as every combination oj members, oj a truss is in a state oj 
rest, the algebraic sum oj the moments oj the jorces acting 071 it, with 
respect to any point in the plane oj the truss, must be equal to zero. 
The forces acting on the member or combination of members may 
be either external jorces directly applied to it or stresses communi¬ 
cated to it through other members. 

It is usually better to take the panel points as the centers 
of moments. Clockwise moments are considered positive; 
stresses acting towards the part of the truss which is considered 
free to move are compressive; those acting away from it are 
tensile. 

In the Howe truss shown in Fig. 45, 



Let 1 = panel length ^ 4 C; 
d = panel depth BC = 


l 


tan p ’ 
b = distance CH = l cos </>. 

Let rotation be assumed about the panel points in succes¬ 
sion. 

























204 


CIVIL ENGINEERING . 


Panel Point C.—Under the action of the force the 

v.' 

member AC would rotate about C were it not restrained by 
the member BA. The moment of the stress BA about the point 
C must therefore be equal and opposite to the moment of the 
force 5IT about the same point, or 


stress BA Xb — $Wl = o; 


stress BA = 


5 ]VI 

b 


jW_ 

COS (f>’ 


(431) 


As its moment is positive, it acts towards the vertex A and is 
a compressive stress. 

Under the action of the force 5IT the triangle ABC would 
rotate about the point C were it not restrained by the member 
DB. Hence the moment of the stress DB about C must be 
equal to the moment of the stress 5IT about the same point, or 


stress BD X d — 5IT/ = o; 


ST VI 

stress BD = = 5 IT tan 


(432) 


As its moment is positive, the stress acts towards B and is com¬ 
pressive. 

Panel Point B.—The reaction 5IT would rotate the member 
AB about B were it not restrained by the member AC; hence 
we have 


stress ACxd — 5IT/ = o; 


.*. stress AC = 


5WI 

d 


= 5 IT tan <f>. 


(433) 


As its moment is positive, this stress acts away from A and is 
tensile. 

Panel Point D.—The reaction 5IT would rotate the broken 
line ABD about the point D were it not restrained by the 
members AC and BC; hence we have 


^ , 5IT/ 7 

stress BCxl+ '-j- Xd-$WX2l=o; 


^ 10WI-AVI 

.’.stress BC= - - =5 IT. . . . 


• • ( 434 ) 








DETERMINATION OF STRESSES BY ANALYTIC METHODS. 205 

As its moment is positive, it acts away from the point B and is 
a tensile stress. 

The parallelogram A BCD is held in a state of rest with 
respect to D by the forces 5 IT, 2 IT, and the stress in the mem¬ 
ber CE\ hence we have 


stress CE Xd — 5 ITX 2/+2IT/=o, 


stress CE 


mi 

d ’ 


-8IT tan </>. 


(435) 


As its moment is positive, it acts away from C and is a tensile 
stress. 

Panel Point E.—The triangle ABC is held in a state of rest 
with respect to the center E by the forces 5IT and 2IT and the 
stresses in BD and CD. Hence we have 


stress BDxd+ stress DCxb — 5ITX 2I+ 2lT/=o. 


_ <Wl 

But stress BD=^—r~ 


hence 


or 


stress DCxb = ioWl—2Wl—$Wl=$Wl> 


stress DC = 


sm 

b 


3IT 

COS 4 ) 


(436) 


As its moment is positive, it acts towards C and is a compres¬ 
sive stress. 

The trapezoid ABDE is held in a state of rest with respect 
to the center E by the forces 5 IT and 2 IF and the stress in the 
member FD. Hence 

stress FDXd—$W X2/ + 2lT/ = o, 

mi 

stress FD=~ = 8IF tan (f> .(437) 


As its moment is positive, it acts towards D and is a compres¬ 
sive stress. 










206 


CIVIL ENGINEERING. 



Howe Truss. 
Fig. 45. 


Panel Point F.—The figure FBACDF is held in a state of 
rest with respect to the center F by the forces 5IF, 2 W, and 
the stresses in CE and DE. Hence 


stress DE X/ + stress CE Xd — 5 W X3/ + 2 IFX2/=o. 


^ 81 VI 
But stress CE =—7—. 

a 


Hence 


stress DE—nWl—ZWl=^Wl, , . . (438) 

As its moment is positive, it acts away from D and is a tensile 
stress. 

The parallelogram FBAE is held in a state of rest with respect 
to the center F by the forces 5 IF, 2 IF, 2 IF, and the stress in 
EG\ hence 


stress EG Xd — 5IF X 3/ + 2 IFX 2I + 2IFX l =o, 


stress EG 


gWl 

d 


9IF tan <f>; . . . 


(439) 


As its moment is positive, it acts away from E and is a tensile 
stress. 

Panel Point G.—The trapezoid DBAE is held in a state of 
rest with respect to the center G by the forces 5IF, 2W, 2IF, 
and the stresses FD and FE. 

Hence 


stress FEXb + stress FDxd-5WX3I+2WX2l+2Wl=o. 


Substituting for FD its value 


mi 

—7- we have 
a 


stress FE F~r = —~t- 

0 COS (p 




. . (440) 











DETERMINATION OF STRESSES BY ANALYTIC METHODS. 207 

As its moment is positive, it acts towards E and is a compres¬ 
sive stress. 

Panel Point D'.— The figure D'BAEFD' is held in a state 
of rest with respect to center D r by the forces 5 PE, 2 PE, and 2 PE, 
and the stresses EG, EG, and E'F. The stress in E'F may be 
assumed as equal to EF on account of the symmetrical loading, 
or may be determined by solving the truss from the point A'. 
Assuming it to be equal to EF we have 

stress FGXl — stress FE r X b + stress GE X d — 5ITX4/ 

+ 2PE X3/ + 2PE X 2/ = o, 

stress FGxl = 20WI + Wl-gWl - 6 Wl - 4WI = 2WI, (441) 
stress FG = 2W. 


As its moment is positive, it acts away from F and is a 
tensile stress. 

The following table gives the stress in each member: 


Members. 

Compressive. 

Tensile. 

BD 

Upper Chord. 

5W tan (J> 


DF 

8W tan (f> 


AC 

Lower Chord. 

5W tan <f) 

EC 


8W tan <j> 

EG 


9W tan $ 

AB 

CD 

Diagonal Web. 

5 W 

COS 

3W 


EF 

BC 

cos 9 

W 

COS cj) 

Vertical Web. 

5 W 

DE 


3 W 

FG 


2W 


Method of Moments in Roof-trusses. —To apply the method 
of moments to the determination of the stresses in the members 























2o8 


CIVIL ENGINEERING. 


of a truss without parallel chords, let it be required to determine 
the stresses in the truss shown in Fig. 46. 


2W 

I 



Fig. 46. 


Let l = AD = length of principal 

rafter or upper 
chord; 

( f> = angle CA F. Then 
/ sin <f> = AE = one-half the span 

AA'; 

l cos <j) = DE = n se of the truss. 


Panel Point C.—To find the stress in AB, take the center of 
moments at C. The reaction 3 IT will rotate the member . 4 C 
about C unless restrained by the member AB. Hence we have 


stress AB X CH — 3 IT X AH = o, 

, „ l COS <j> TTr l sin <f) 
stress A B X-= 3 IT X- 


stress AB=t ) W tan (ft .(442) 


As its moment is positive, the stress acts away from A and is a 
tensile stress. 

Panel Point D.—To find the stress in BB', take the center 
of moments at D. The reaction 3 IT and the force 2 IT will rotate 
the triangle ABD about D unless restrained by the member BB'. 
Hence we have 


stress BB'XDE — ^WXAE + 2WXHE = o, 

stress BB'Xl cos </> = 3l VXl sin 6 — 2TTX ~—- 

2 

= 2 IT/ sin cj), 

stress BB' = 2IT tan </> .(443) 

As its moment is positive, the stress acts away from B and is a 
tensile stress. 

Panel Point B.—To find the stress in ^ 4 C, take the center 
of moments at B. The reaction 3IT will rotate the member AB 
about B unless restrained by the member CA. Hence we have 












DETERMINATION OF STRESSES BY ANALYTIC METHODS. 209 


stress CA XBC - $WxAB = o, 


BC = 


l 

2 tan 0’ 


AB = 


l 

2 sin 0’ 


stress CA X- 7 

2 tan 0 


jWl 
2 sin 0’ 


stress CA = 


3 W 

COS 4 > 


- • (444) 


Since its moment is positive, the stress acts towards A and is 
a compressive stress. 

To find the stress in DC , take the center of moments at B. 
The reaction 3 W and the force 2IT will rotate the triangle ABC 
about B unless restrained by the member DC. Hence we have 


stress DCxBC-2 > WxABA2WxBH = o j 


BH = BC cos 0 = 


1 COS 4 > 

2 tan 4> 


1 cos 2 4> 

2 sin 0 ’ 


1 cos 4> 
stress DC X — X -— 7 

2 sin 9 


3IT/ 2IT/ cos 2 4 > 

2 sin 4> 2 sin 4> 9 


stress DC = 


3 ^ 

COS 4 > 


— 2 IT cos 



(445) 


As its moment is positive, this stress acts towards C and is a 
compressive stress. 

Panel Point D.—To find the stress CB, take the center of 
moments at D. The reaction 3IT and the force 2 IT will rotate 
the member ACD about D unless restrained by the members 
BC and AB. Hence we have 


stress BCxCD — $WXAE + 2WXHE + stress ABxDE = o, 

1 . 2WI sin 0 

stress BCX — = ^WlXsm 0 —-— 3W tan 0 X / cos 0 

2 2 

= - IT/ sin 0 , 


stress BC= — 2IT sin 0 


( 446 ) 
















2 10 


CIVIL ENGINEERING. 


As its moment is negative, this stress acts towards C and is a 
compressive stress. 

Panel Point C.—To find the stress BD , take the center of 
moments at C. The reaction will rotate the triangle ABC about 

2W 

I 



C unless restrained by the members BB ' and BD. Hence we have 
stress BDxCI — 3 WXAH + stress BB'XCH = o, 

l cos d> 

IC = CH =-S 

2 

/ cos A AVI sin 6 , / cos d> IT/ sin 0 

stress BD X -=-— 2W tan d> X -=-, 

2 2 22 

stress BD = W tan p .(447) 


As its moment is positive, this stress acts away from B and is 
a tensile stress. 

The stresses in all the members may be tabulated thus: 


Members. 

Compressive. 

Tensile. 

AC 

3 W 




COS (j) 



CD 

■— 2W cos d> 



cos 9 



BC 

2W sin <jS 



AB 

3 \v 

tan <}> 

BB' 


2 W 

tan <f> 

BD 


w 

tan <f> 


Method of Sections.—The method of moments is often 
called the method of sections, since, if we pass a plane through 
a truss cutting its members, the algebraic sum 0) the moments oj 
the stresses in the pieces cut , and oj the external jorces which act 
on the truss on one side oj the plane about any point in the plane 
oj the truss , will be equal to zero. The resulting equation may be 
solved when it contains but one unknown intensity. This will 
be the case when only two members are cut and the center oj 


























DETERMINATION OF STRESSES BY ANALYTIC METHODS. 211 


moments is taken on one oj the members; when three members which 
are not all parallel to each other are cut and the center oj moments 
is taken at the intersection oj two oj them; when the intensities oj 
all the members cut , save one , are known. 

Thus in Fig. 47 we may determine the stresses in all the 
members of the panel FDGE 
in the following manner: 


Let l = AC\ 
d = BC = 


l 


tan p ’ 
b = CH = / cos (j>. 



Howe Truss. 
Fig. 47. 


Pass a plane cutting the members FD, FE, and GE ; let F, 
the intersection of the members FD and FE, be the assumed 
center of moments. If we consider the part of the truss A'B'FGA 
to the left of the section, to be at rest, the part ABDE , to the 
right of the section, would rotate about F due to the extraneous 
forces acting at A , C, and E were it not restrained by the stress in 
GE. Hence we have 


stress GEXd— 5 IFX3/+2 W X 2/ + 2WI = o, 


_ 9 Wl _ 
stress GE=—— = 9 IV tan <p. 


( 448 ) 


As the moment of the stress GE is positive, it must act away from 
E and be a tensile stress. 

If E, the intersection of FE and GE, is taken as the center of 
moments, we have 

stress FDxd—$WX2l+2\Vl=o i 

81 VI 

.*. stress FD = —p = 8IF tan <p .(449) 


As its moment is positive, it must act towards D and be a com¬ 
pressive stress. 

If we take the center of moments at G, we have 
stress FExb + stress FD X d - 5IFX3 /■+ 2TFX 2/ + 2 Wl =o; 















212 

CIVIL ENGINEERING. 


but 

FDxd=SWl, hence 



stress FEXb = gWl — SWl = Wl, 



IT 7 

stress FE — ........ 

cos <p 



As its moment is positive, the stress must act towards E and 
be compressive. 

Pass a plane cutting E'G, FG, FE, and FD, and take the 
center of moments at E'. Then 

stress FGXl + stress FE X2 b + FDXd — 5TTX4/ + 2IT 7 X3/ 

+ 2 W X 2I -f 2WI =0, 

stress FGxl= - 2WI-SWl +20JJ 7 /-6 Wl- 4 Wl- 2 Wl, 
stress FG = — 2W .(45 r) 

Since its moment is negative, it must act away from G and be a 
tensile stress. 

Pass a plane cutting EC, DE , and FD, and take C as the 
center of moments. Then 

stress FD X l + FD Xd — 5 Wl=o, 

stress EDxl=$Wl-SWl = -3JIT, 

stress FD= — 3 IT 7 .(452) 

Since its moment is negative, it must act away from D and be a 
tensile stress. 



Howe Truss. 
Fig. 47. 













CHAPTER XI. 


EFFECT OF MOVING LOADS UPON TRUSSES WITH 

PARALLEL CHORDS. 

Since a truss with parallel chords is simply a beam in which 
the chords must resist the bending moment and the web members 
must resist the shear, the same principles must govern them 
both. 

Concentrated Live Load.—As applied to trusses proposition 
I, page 119, should read: If a single concentrated load is moved 
over a truss without weight resting on end supports, the stress in 



any chord member will be numerically greatest when the load is 
between the middle point oj the member and the middle point oj 
the truss , and as near to the former point as the method oj loading 
will permit. Thus in the Warren truss, Fig. 48, if it is assumed 
that the load can be placed on either the upper or lower chord, 
the greatest stress in the member FD is produced by placing the 
load at E, and the greatest stress in the member GE is produced 
by placing the load at F. 

If the load is confined to the panel points of a single chord, 
as the lower, the greatest stress in a member of that chord is 
produced by placing the concentrated load at that end of the 
member which is nearest the middle point of the truss. Thus 
if the load is applied at G, it will produce a greater stress in GE 
than if placed at E. 

As applied to a truss, proposition II, page 121, should read: 
The chord member having the maximum stress is the one at the 
middle point oj the truss. 


213 




214 


CIVIL ENGINEERING. 


The maximum chord stress is in the middle member of the 
upper chord FF' when the concentrated load is at G; the maxi¬ 
mum chord stress in the lower chord is in the middle members 
E'G and EG also when the load is at G. 

Proposition III, page 121 should read: Propositions I and II 
are equally true ij the truss is oj uniform weight per lineal foot 
and its weight is considered , or if we combine the live with a uni¬ 
form dead load. 

Proposition VIII, page 129, should read: If a concentrated 
load is moved over a truss without weight resting on end supports, 
the stress in any web member will be numerically greatest when 
the load is between the middle point of the truss and the 
member , and as near the member as the method of loading will 
permit. 

Thus in Fig. 48 the stress in DC will be a maximum when 
the concentrated load is at E. 

The final conclusion on page 130, under the same propo¬ 
sition, should read: The maximum stress in the web members of 
the end panels is greater numerically than the maximum stress 
in any other web members. 

Thus in Fig. 48 the maximum stress in the web members of 
the panels ABC and A'B'C' is greater than the maximum stress 
in any other web members. 

From equations (350) and (351) it follows that the shear on 
one side of the concentrated load is positive and on the other 
negative. In a truss this must be interpreted as showing that 
in any system of parallel web members , those on one side of the 
load will be under tension , and those on the other under com¬ 
pression. 

Thus, in the truss Fig. 48, if the concentrated load is at £, 
the stress in the members ED and BC is tension, but the stress 
in the parallel members FG, F'ED'C ', and A'B' is com¬ 
pression. 

Proposition IX, page 130, should read: If a concentrated 
load moves over a truss of uniform weight, the live- and dead-load 
web stresses will have the same signs between the load and the 
nearer support , as well as between the middle point of the beam 
and the farther support , but will have unlike signs between the load 
and the middle point. 


LOADS UPON TRUSSES. 


215 


Thus, in the truss Fig. 48, if the truss has a concentrated load 
at E and a uniform dead load, the stress in any member between 
E and A or between G and A' will be the numerical sum of the 
live- and dead-load stresses; but in the members EF and FG the 
resultant stress is the numerical difference between the stresses 
due to the live and dead loads. 

It follows that there will be a point oj ?io shear in the truss as 
in the beam , and if this point is at least one panel length from the 
middle point , the stress in some oj the web members will be re¬ 
versed as the live load crosses the truss. 

As in beams, page 132, the resultant stress in any web member 
is greatest when its live-load component stress is greatest. 

Uniformly Distributed Live Loads. — Proposition V should 
read: If a live load of uniform weight moves over a truss without 
weight, resting on end supports separated by a distance less than 
the length of the live load, the stress in any chord member will 
be greatest when the live load entirely covers the truss. 

Proposition VI should read: Proposition V is equally true if 
the truss upon which the load is moved is oj uniform weight and 
its weight is considered , or if we combine the live and dead loads. 

Proposition X should read: If a uniformly distributed live 
load is moved over a truss of uniform weight resting on end 
supports separated by a distance less than the length of the 
load, the stress in any web member will be algebraically greatest 
when the live load covers the greater segment into which its panel 
divides the truss , and algebraically least when the live load covers 
the smaller segment. 

To illustrate these statements, let the Warren truss, shown 
in Fig. 48, be subjected to a concentrated load 6 W which acts 
in succession at each panel point of the lower chord. Determine 
the numerical coefficients by any of the methods heretofore 
explained, and tabulate them in columns 1 to 6 as shown. To 
determine the actual stresses, chord coefficients must be multi¬ 
plied by tan <p, and web coefficients by 



Fig. 48. 






2 l6 


CIVIL ENGINEERING. 


I 

2 

3 

4 

5 

6 

7 

Moving Load of 6 W at 


C 

E 

G 

E' 

C' 

Uniform Dis¬ 
tributed Load 
of 30 W. 


Upper Chord. 


BD 

— 10W 

- 8W 

- 6W 

- 4W 

- 2W 

. - 30W 

DF 

- 8W 

-16W 

-12W 

- 8W 

- 4W 

-48W 

FF' 

- 6W 

-12W 

-18W 

-12W 

- 6W 

-54W 

F'D' 

- 4W 

- 8W 

-12W 

-16W 

- 8W 

-48W 

D'B' 

- 2W 

- 4 W 

- 6W 

- 8W 

— 10W 

-30W 


Uower Chord. 


AC 

+ 

5 W 

+ 4W 

+ 

3 W 

+ 

2W 

+ 

iW 

CE 

+ 

9W 

+ I2W 

+ 

9W 

+ 

6W 

+ 

3 W 

EG 

+ 

7W 

+ I4W 

+ 

I5W 

+ 

10W 

+ 

5W 

GE' 

+ 

5 W 

+ IOW 

+ 

I5W 

+ 

14W 

+ 

7W 

E'C' 

+ 

3 W 

+ 6W 

+ 

9 W 

+ 

12W 

+ 

9 W 

C'A' 

+ 

iW 

+ 2W 

+ 

3 W 

+ 

4 W 

+ 

5W 


Web Members Parallel to AB. 


AB 

— 

5 W 

— 

4 W 

— 

3 W 

— 

2W 

- iW 

CD 

+ 

iW 

— 

4W 

— 

3 W 

— 

2W 

- iW 

EF 

+ 

iW 

+ 

2W 

— 

3 W 

— 

2W 

- iW 

GF' 

+ 

iW 

+ 

2W 

+ 

3 W 

— 

2W 

- iW 

E'D' 

+ 

iW 

+ 

2W 

+ 

3W 

+ 

4 W 

- iW 

C'B' 

+ 

iW 

+ 

2W 

+ 

3 W 

+ 

4 W 

+ 5W 


+ 15W 
+ 39 W 
+ 51W 
+ 51W 
+ 39W 

+ 15W 


-15W 

- 9W 

- 3W 
+ 3W 
+ 9W 
+ 15W 


Web Members Parallel to A'B'. 


BC 

+ 5 W 

+ 

4 W 

+ 

3W 

+ 

2W 

+ 

iW 

-f 

I5W 

DE 

- iW 

+ 

4 W 

+ 

3W 

+ 

2W 

+ 

iW 

+ 

9W 

h G 

- iW 

— 

2W 

+ 

3 W 

+ 

2W 

+ 

iW 

+ 

3W 

F'E' 

- iW 

— 

2W 

— 


+ 

2W 

+ 

iW 


3W 

D'C' 

- iW 

— 

2W 

— 

3W 

— 

4W 

+ 

iW 

— 

9W 

B'A' 

— i w 


2W 


3 W 

■ 

4W 


5 W 

— 

I5W 


Exercise. —Construct a similar table for a five-panel Warren 
truss whose panel load is 5 W. 


Concentrated Load Alone.—If this table is examined, it will be 

seen that, as stated above, the greatest stress occurs in any given 
member 0j the upper chord when the concentrated load acts at the 
panel point beneath it. Thus the greatest stress, — 10W tan p } 
occurs in the member BD when the concentrated load is at C. 

The greatest stress occurs in any given member oj the lower 
chord when the concentrated load acts at the adjacent panel 
point nearest the middle oj the truss 1 thus the greatest stress, 
+ 12 W tan p, occurs in CE when the concentrated load is at E . 












































LOADS UPON TRUSSES. 


217 

The maximum stress in either chord occurs in the middle 
member or members when the load is at the middle panel point. 
Thus the maximum stress in the upper chord is -1 8W tan <p 
in the member FF', and in the lower chord +15 W tan <p in 
EG and E'G, when the concentrated load is at G. 

In any web member the stress is a maximum when the con¬ 
centrated load is between the member and the middle point oj the 
truss and as near to the member as the method oj loading will per- 

mit. Thus the greatest stress in FE is — -— when the concen- 

cos 

trated load is at G. It is a compressive stress. 

In any web member the stress is a minimum when the concen¬ 
trated load is between the member and the nearer support and as 
near to the member as the method oj loading will permit. Thus 

2 W 

the minimum stress in EF is + — when the load is at E. It is 

cos 

a tensile stress. 

By the maximum stress in any member is meant the greatest 
stress which is of the same character as that produced by the dead 
load. 

If the stress in the member is not reversed, its minimum stress 
is the least stress of the same character as the maximum stress; 
if the stress in the member is reversed, the minimum stress is 
the greatest stress of an opposite character jrom the maximum. 

The maximum web stresses are — 5IT in AB and B'A' and 
+ 5IT in C'B and BC. 

In a series oj parallel web members the stresses are all nega¬ 
tive , or compressive , on one side oj the concentrated load and posi¬ 
tive , or tensile, on the other side oj the load. Thus when the con¬ 
centrated load acts at E the stress in AB and in CD is negative, 
and in EF, GF', E'D ', and C'B' positive. 

Dead and Concentrated Moving Loads Combined.—If we 
add together the coefficients given for each member in the col¬ 
umns 2, 3, 4, 5, and 6, the result, given in column 7, will be the 
coefficients in all the members due to a uniform dead load whose 
panel length weighs 6 W. 

If we now assume that members of the truss are subjected to 
this dead load and at the same time are subjected to a moving 
load as before, we may determine the combined effects of the 




CIVIL ENGINEERING. 


218 


single concentrated load and the uniform dead load by adding 
the stresses taken from column 7 to those given in columns 
2, 3, 4, 5, or 6, as the concentrated load moves from A to C, E, 
G, E', and C'. 

The greatest stress occurs in any given member 0j the upper 
chord when the concentrated load is at the payiel point beneath 
it; thus the stress in DF is a maximum, — 64IT tan 0 , when the 
concentrated load is at E. 

The greatest stress occurs in any given member oj the lower 
chord when the concentrated load is at the adjacent panel point 
toward the center; thus the stress in EG is a maximum, 
+ 66IT tan 0, when the concentrated load is at G. 

The maximum stress in either chord is in the middle member 
when the concentrated load is at the middle panel point. Thus 
the maximum chord stress is —72IT tan <p in the member FF' 
when the concentrated load is at G. 

It will be observed that the stress in every member oj the upper 
chord is compression and in the lower chord tension , whatever 
be the method oj loading. 

In any web member the stress is a maximum when the concen¬ 
trated load is between the member and the middle point oj the truss 
and as near to the member as the method oj loading will permit; 
the stress is a minimum when the load is between the middle point 
and the nearer support and as near to the member as the method 
oj loading will permit. Thus the stress in EF is a maximum, 


6IT . W 

—-when the load is at G, and a minimum,- r , when 

COS 0 COS 0 

the load is at E. 

As these stresses are both compressive there can be no 
reversal of stress in the web members if the panel weight of 
the dead load is equal to that of the moving load, as we have 
assumed. 

The web members oj the panels on either side oj the middle 
point , which have the least stress under the dead load alone , are the 
ones in which the stresses will be the first reversed; therefore, if 
we find that the maximum and minimum stresses in FG and 
GF' are of the same character, we may conclude that the rela¬ 
tive weights of the concentrated and dead loads are such that the 
limiting points of zero shear do not move as far from G as E 




i 


LOADS UPON TRUSSES. 


219 


or E'. From the table we see that the maximum stress in FG 

a - , 6W 1 4l . W 

and G -b is - 7, and the minimum is +- 7. 

COS <p COS ( f > 

To find the value of concentrated load which placed at E r 
will reduce the stress in GF' to zero, we shall have, from 
previous principles, 

R 1 ~W — 2 W = o, or 2 \w + §IF — W — 2 W = o, 

in which W = weight of moving load; 

w = ‘‘ “ panel length of dead load; 

Ri = reaction at A'. 

Hence W = fw, placed at E', will reduce the stress in GF' and. 
E'F' to zero. A greater value of W would reverse the stress 
in these members. In a similar manner we may determine the 
value of W which, placed at C', will reverse the stress in GF', 
and the value of W which will reverse the stress in F'D' and D'C '. 

The maximum stress in the web members occurs in the end 
panels when the co 7 icentrated load rests at the adjacent panel point; 

20IT 

thus the maximum stress is —- 7, which is the stress in AB and 


+ 


20W 

COS <f> 


COS </>’ 

in BC when the load is at C. 


Moving Load of Uniform Weight per Unit of Length, whose 
Length Equals that of the Truss.—The effect of such a load upon 
a truss is similar to its effect on a simple beam. 

Whether the live load acts alone or in conjunction with the 
uniformly distributed weight of the truss, the stress in any member 
of either chord is a maximum when the moving load covers 
the entire truss; it is a mininum when the moving load is entirely 
off the truss. Thus, in Fig. 48, if the load is confined to the lower 
chord, the maximum stress in CE occurs when the moving load 
acts at the panel points C, F, G, E' , and C'\ it is a minimum when 
it acts at none of these panel points. 

The maximum chord stress occurs in the middle members when 
the load covers the entire truss: in FF' of the upper chord, and in 
GE and GE' of the lower chord. 

The stress in any web member is a maximum when only the 
panel points of the longer segment into which the member divides 
the truss are acted upon by the moving load, and a minimum when 






220 


CIVIL ENGINEERING. 


only the panel points oj the shorter segment are acted upon. Thus, 
in Fig. 48, the stress in GF is a maximum when the moving 
load acts only at C', E', and G, and a minimum when it acts 
only at E and C. 

The maximum web stress oj the truss occurs in the members 
oj the end panels when all the panel points are covered. 

To illustrate these statements, if in the table above given 
we add to the stress given in column 7 the stresses given in 2, in 
2 and 3, in 2, 3, and 4, in 2, 3, 4, and 5, in 2, 3, 4, 5, and 6, 
in 3, 4, 5, and 6, in 4, 5, and 6, in 5 and 6, and in 6, we may 
determine the effect of a moving load whose panel length weighs 
6 W upon a truss whose panel length also weighs 6 W. 

The maximum stress in the chord member CE , + 78IT tan p, 
is obtained by adding the stresses due to loads of 6IT at C, E f 
G, E ', and C' to the stress in that member due to the dead load. 
The minimum stress in that member is + 39JT tan p, due to the 
dead load alone. The maximum chord stress is —108IF tan p in 
FF' of the upper chord, when all the panel points are loaded; 
the maximum stress in the lower chord is +102 IT tan p in EG 
and GE'. 

The maximum stress in the web member EF occurs when the 
panel points in the longer segment, C', E ', and G, are loaded; 
qW 

and is-7. The minimum stress o in the web member EF 

cos 9 

occurs when both the panel points C and E, in the shorter segment, 
are loaded. 


The maximum stress in the web members is the stress ~^ 

cos p’ 

which occurs in the members of the end panels when the truss 
is fully loaded. 

To determine whether there is any reversal oj stress in the 
web members , we first determine the maximum and minimum 
stresses in the middle members GF and GF'. The maximum 

Q W 

stress in these members is +--, and the minimum stress 

COS o 

zero; there is therefore no reversal when the panel lengths of the 
dead and moving loads are of equal weight, as we have assumed. 

Any Increase in the weight of the moving load would, how¬ 
ever, reverse the stresses in GF and GF'. 





LOADS UPON TRUSSES. 


221 


Counterbraces.—It is evident, from the discussion above 
given, that the stress in any web member of a truss, due to a 
uniform dead load, may be reversed or changed from a tensile 
to a compressive stress or the converse, by placing an additional 
load upon one or more of the panel points between the web mem¬ 
ber and the nearer support. The intensity oj this load must 
be such that its reaction at the farther support will be greater 
than the vertical component of the stress due to the dead load alone 
in the member considered. 

If the live load is a concentrated one, the least intensity of such 
a load which will cause reversal of stress in any web member is 
ascertained by placing it at the adjacent panel point on the 
side of the nearer support and making its reaction at the farther 
support equal to the numerical coefficient of the dead-load stress 
in the member considered. The least intensity of a concen¬ 
trated load which will produce a reversal of stress in E'F', Fig. 48, 
is the load placed at E', whose reaction at A is equal to the nu¬ 
merical coefficient of the dead-load stress in E'F'. A concentrated 
load at C' may also reverse the stress in E'F', but its weight 
must be greater than that of the load placed at E'. 

If the live load is a distributed one, its least weight per panel 
length which will reverse the dead-load stress in any member 
is determined by placing it so as to cover all the panel points 
between the member and the nearer support. In this position 
its reaction at the farther support must be greater than the 
numerical coefficient of the dead-load stress in the given member. 
The least weight per panel length of the distributed live load 
which will produce a reversal of the dead-load stress in E'F' may 
be ascertained by placing the load at C and E' simultaneously. 

The reversal of stress in web members of bridge-trusses is 
produced by the passing of heavy wagons and trains. To pro¬ 
vide for this reversal, the truss must be counterbraced ; that 
is, the web members which are subject to reversed stress must 
either be so arranged that they can safely bear the greatest ten¬ 
sile and compressive stresses to which they may be subjected, 
or, in addition to the main braces , which are the web members 
designed to carry the dead load, there must be additional mem¬ 
bers or braces to carry the reversed stresses. These are called 
counterbraces. 


222 


CIVIL ENGINEERING. 


In a Warren truss the counterbracing consists in making 
the web members themselves of such forms and materials as to 
resist both compression and tension, and in connecting them with 
the chords by joints which will also resist tension and compres¬ 
sion. In a metal Warren truss the web members are usually 
angles riveted to the chords. 

In the Pratt, Howe, and similar trusses with vertical web 
members the counterbraces are diagonal members crossing 
the main braces. In Fig. 43 the counterbraces would be tie- 
rods connecting H and E, F and C, etc., and in Fig. 45 they 
would be struts connecting D and G, B and E , etc. 

By this method of counterbracing each web member of 
the truss is subjected only to a stress of compression or to one 
of tension. This makes it possible to design a simpler and 
more economical truss than one in which the web members are 
subject to reversed stresses. 

If the live load is a concentrated one, the maximum reversed 
or counterbrace stress in any web member is ascertained by 
placing the concentrated load at the adjacent panel point on 
the side of the nearer support. 

If the live load is a distributed one, the maximum counter¬ 
brace stress is ascertained in a similar manner by placing the 
live load on all the panel points between the member and the 
nearer support. 

The counterbraces must be placed in all panels in which 
reversal oj stress in the web members may occur under the heaviest 
moving load jor which the truss is designed. They are strongest 
in the middle panels oj the truss and decrease in strength towards 
the supports. 

Queen-post Truss.—The function of counterbraces may be seen 
from a discussion of the simple queen-post truss shown in Fig. 49, in 

which the angle between AB and the 
vertical is assumed to be <p. 

The verticals BC and B'C', although 
a ties, are called queen-posts. If this truss 
Fig 49 is uniformly loaded on its lower chord, 

that is equal weights, as 3IF, are attached 
atC and C', the truss will be in equilibrium and will support these 
weights without the diagonals B'C and BC'. The stress in AB 







LOADS UPON TRUSSES. 


223 


tW 

and A'B' will be in AC, CC' , and AC, + 3IV tan </>; 

in BC and B'C' , + 3IF; and in 55 ', — 3 IF tan </>. There is 
therefore equilibrium at every point. 

Now conceive the joints to be pin-joints, and the truss to 
be acted upon by the single weight 3 IF at C; the reactions would 
be 2W at A, and IF at A', since A'C' =C'C = AC. As it is 
impossible to have equilibrium at every panel point under the 
action of these three forces, the truss will be deformed by the 
rectangle B'BCC' becoming a rhomboid, and finally the top 
chord will fall out. The weight will then be transferred to the 
lower chord and the weight will be supported by a catenary 
instead of a truss, if the end pins at A and A' are firmly fixed. 

If we place a tie-rod from C to 5 ' or a strut from 5 to C', 
the movement of the panel points of the truss will be prevented. 
In either case the reaction at A will be 2IF, and at A', W. 

If the tie C 5 ' is used, the stresses in the different member of 
the truss will be 


AB = 


2W 

COS 4 


AC= + 2IF tan <j), 


BC=+ 2IF, 


55 ' = —2IV tan <£, 


A'B' = 


W 

COS (p 


B'C' = 0, 


B'C = +* 


W 

COS cj>’ 


A'C' = + IV tan <f>, 
CC' = + W tan (f >, 


and there is equilibrium at every point. 

If a strut BC' had been used instead of the tie B'C, the stresses 

would have been 


AB= — 


2W 

cos </>’ 


AC= +2W tan (f>, 


A'B' = — 


W 

cos </>’ 


B'C' = A W, 


55 ' = —W tan 
BC= +3W, 


A'C'=+W tan & 

T W 
BC' = 


cos </> ’ 

CC' = +2 IT tan 0 , 

and there is also equilibrium at every panel point. 









CIFIL ENGINEERING. 


224 


Truss Loading.—In the conventional method of determining 
the stresses due to a moving load as above described, we assume 
that one panel point bears its maximum load, while the next is 
unloaded. This would not be true in practice; the load on each 
panel point could only be a maximum when the load extended 
to the panel points on either side. If the maximum and mini¬ 
mum stresses in the web members are desired with great accu¬ 
racy, the stresses due to the true loading of the truss must be 
determined instead of the stresses due to the conventional load¬ 
ing above described. However, as the conventional method 
of loading produces slightly greater maximum and minimum 
stresses, its errors are on the side of safety, and, being simpler, it 
is the loading usually assumed. 

In determining the maximum and minimum stresses in a 
truss due to a moving load, it will be observed that we need only 
determine and tabulate the chord stresses when the moving load 
covers the entire truss and when it is entirely off the truss. We 
must determine the stresses in each web member when the longer 
and the shorter segments into which it divides the truss are cov¬ 
ered by the live load; and we must place counterbraces in every 
panel containing a web member subject to reversed stress. If 
the counterbrace makes the same angle with the vertical as the 
main brace, the amount of stress in it will be equal to the 
amount of reversed stress in the main brace which it replaces. 

PROBLEMS. 

50. In the Pratt truss, Fig. 43, what is the least concentrated 

load which, placed at C, will reverse the stress in FG ? What is 
the least concentrated load which, placed at E, will produce the 
same result? Ans. 6 IT and 3IT. 

51. What is the weight per panel length of the least uniform 

load which, moving over the Pratt truss, Fig. 43, will reverse the 
stress in FG? Ans. 2IT+. 

52. A concentrated load of 4ITis moved over the lower chord of 
the Howe truss, Fig. 45. Insert counterbraces where necessary in 

IT 

the truss, and find counterbrace stresses. Ans. GD and GD '; —. 

3 

53. A uniformly distributed live load whose weight per panel 
length is 3IT moves over the same truss; insert the necessary 
counterbraces and determine the stresses in them. 


IT 


CHAPTER XII. 


GRAPHIC METHOD OF DETERMINING THE STRESSES 

IN A TRUSS. 

The Simple Truss Problem.—In the solution of a simple 
truss by the graphic method, the same problems arise as in its 
solution by the analytic methods. The system of forces or 
stresses acting at each panel point is a concurrent coplanar 
system in equilibrium, of which the line of direction and a point 
of application of each force are given. Such a system can be 
solved graphically as well as analytically when the intensities of 
all the forces except two are known. The system of forces act¬ 
ing upon the truss as a whole is usually a non-concurrent system, 
in which all the forces are known save the two reactions; of 
the reactions, the points of application are known, and also one 
or both of the action lines. Such a system can be solved graphi¬ 
cally as well as analytically when the intensities of the reactions 
and one of the action lines are unknown. 

Representation of Forces. — In the graphic method the 
intensity of a force in terms of the unit of force is not expressed 
in figures, as in the analytic method, but it is expressed by the 
length of a right line drawn to a scale whose unit of length repre¬ 
sents a unit of force. By means of the scale, the intensity of 
a force given in figures may be expressed in graphic units, or 
the converse. 

The direction of a force is indicated by an arrow-head written 
on its action line; if the force acts towards a material point, 
the arrow-head is directed towards the point; if the force acts 
away from the point, the arrow-head is directed away from the 
point. 

Concurrent System—Problem I. —To find the intensity and 
action line oj the resultant oj any concurrent system oj coplanar 
jorces. 


225 


226 


CIVIL ENGINEERING. 


In Fig. 50 let AB, BC, and CD be the action lines of three 
forces forming a concurrent system whose resultant is desiied. 

Let the intensity of each force be 
represented by the length of its 
, action line. Let the unit of length 
of the scale represent an intensity 
of one pound. 

From any point, as A', draw a 
line A'B' parallel to AB and make 
it equal to ten linear units; from 
B f draw a line B'C' parallel to 
BC and make it equal to thirteen 
linear units; from C draw the 
line C'D' parallel to CD and 
make it equal to eight linear units; complete the polygon by 
drawing the line A'D'. If through the common vertex we draw 
a line parallel to A'D', it will be the action line of the resultant 
of AB, BC, and CD, and its intensity will be the number of linear 
units in the length A'D'. This results from the principle of 
the parallelogram of forces, since by construction A'C' must be 
equal and parallel to the resultant of AB and BC, and A'D' 
must be equal and parallel to the resultant of A'C' and CD, 
or of the three forces given. 

Graphic Nomenclature. — In Fig. 50 the diagram A BCD 
shows the actual positions as well as the directions of the known 
forces in the plane of construction; it is therefore called a posi¬ 
tion diagram. In this diagram the forces are indicated by the 
letters written between the action lines. 

The diagram A'B'C'D' is called the jorce polygon, and 
always gives the intensities of the forces it represents. It will 
also give the directions of the forces, if care is taken to draw all 
the known forces in the directions in which they act; thus the 
force AB acts in the direction A'B'. Each force in the force 
polygon is indicated by the same letters as in the position dia¬ 
gram, but in the force polygon the letters are written at the ends 
of the forces and are distinguished by accents. In constructing 
the force polygon A'B'C'D' the forces were drawn in the direc¬ 
tion of the arrow-heads, and A'D' was found to be the direction 
and intensity of the resultant. As a similar result would have 


"^ \\\ / 

/ E 1 
! % Xv \ \ 1 






ili'ii ii.it 


10 15 20. lbs. 


Fig. 50. 



DETERMINING THE STRESSES IN A TRUSS. 


227 


been obtained in solving any other system of concurrent forces, 
we may conclude that: 

// the forces in the force polygon are drawn in the direction 
in which they act, the closing line, drawn from the origin to the 

extremity of the last force, will represent the intensity and direction 
of the resultant of the forces. 

Problem II. To find the unknown intensities of two forces 
of a concurrent system in equilibrium when the action lines of 
all the forces, and the intensities of all the other forces, are known. 

In Fig. 50 let AB, BC, and CD be the action lines of the 
known forces, and DE and EA be the action lines of the un¬ 
known forces whose intensities are desired. As before, construct 
the force polygon A'B'C'D from D' draw the line D'E' par¬ 
allel to DE, and from A' the line A'E' parallel to AE. Then 
the number of linear units in D'E' and E'A' will be the inten¬ 
sities in pounds of the unknown forces. According to Problem I, 
A'D' is equal and parallel to the resultant of AB, BC, and CD, 
and D'A' is equal and parallel to the resultant of DE and EA. 
Since A'D' and D'A' are equal and contrary, the five forces 
AB, BC, CD, DE, and EA must form a concurrent system in 
equilibrium, and D'E' and E'A ' must be the intensities of the 
unknown forces. 

Other Uses of the Force Polygon.—The force polygon may 
be utilized to determine whether a system of concurrent forces 
is in equilibrium or has a resultant. If the force polygon closes, 
the sum of the horizontal and the sum of the vertical compo¬ 
nents of the system of forces must each be equal to zero, and 
the concurrent system must be in equilibrium. Conversely, if 
the concurrent system of forces is in equilibrium, its force poly¬ 
gon must close. 

Any force of the system may be resolved into two components 
acting along any desired lines of direction. Thus the force DA, 
Fig. 50, if drawn through the common vertex, may be resolved 
into two components parallel to DE and AE, by simply draw¬ 
ing through the extremities of D'A' the lines D'E' and A'E', 
parallel respectively to DE and AE. 

The force polygon may be utilized in determining two unknown 
directions in a concurrent system of forces when all the inten¬ 
sities are known. If the intensities instead of the directions of 


228 


CIVIL ENGINEERING. 


AE and ED were given, by describing arcs with A' and D' as 
centers and the intensities as radii the point E' could be deter¬ 
mined. In the general case the arcs will intersect in two points 
and give two solutions of the problem. 

The force polygon may also be utilized in determining one 
unknown intensity and one unknown action line. Thus if the 
action line of DE and the intensity of AE were given, the point 
E' could be determined by drawing D'E' parallel to DE , and 
from A' as a center with the intensity of EA as a radius describ¬ 
ing an arc intersecting D'E'. Under these conditions there 
would also be two solutions in the general case. If both the 
intensity and the action line of either DE or EA were given, 
there would be but one solution. 

Non-concurrent Forces—Problem III.— To determine the re¬ 
sultant oj a system oj non-concurrent coplanar forces. 

In Fig. 51 let AB, BC, and CD be the action lines of the 
three forces whose intensities are given by the force polygon 
A'B'C'D'\ required the resultant. 



The direction and intensity of the resultant is given by the 
closing line of the force polygon A'D'. This results as before 
from the principle of the parallelogram of forces, since by con¬ 
struction A'C' must be equal and parallel to the resultant of 
AB and BC , and A'D' must be equal and parallel to the result¬ 
ant of ^ 4 'C' and CD. 



DETERMINING THE STRESSES IN A TRUSS. 


229 


The resultant will be fully known, therefore, whenever we 
know the position of one point of its action line. 

When the action lines of the forces intersect within the limits 
of the drawing we may employ the following method based on 
Problem I. Produce the action lines of AB and BC until they 
intersect. Through the point of intersection draw their result¬ 
ant which will be parallel to A'C'. Produce this resultant 
until it intersects the action line of CD ; their point of inter¬ 
section will be one point on the action line of the resultant, and 
the resultant itself will be AD, drawn through this point parallel 
to A'D', as shown in Fig. 51, diagram I. 

When the action lines of the forces do not intersect within 
the limits of the drawing we must employ the following method: 

Assume a point O, Diagram II, called the pole of the force 
polygon, at any convenient point in the construction plane, and 
draw the lines A'O, B'O, C' 0 , and D'O. These lines may be 
taken as the intensities and directions of a new system of forces 
which are components of the original system. A'O and OB' will 
be the intensities and directions of the components of AB, and 
B'O and OC' will be the intensities and directions of the com¬ 
ponents of BC, etc. 

If we assume any point on the action line of the force AB, 
as o', and draw the lines o'a and o'b equal and parallel to A'O and 
OB', the original force AB may be replaced by its components 
o'a and o'b. If we find the point 0" where the line bo' intersects 
the action line of the force BC, we may replace the force BC 
by its components o"b and o"c, equal and parallel to B'O and 
OC'. If we find the point o'" where the line o"c intersects the 
action line of the force CD, we may also replace the force CD 
by its components o'"c and o'"d, equal and parallel to C'O and 
OD' . The resultant of these six components, or the resultant 
of the three resultants arising from taking these components in 
pairs, must also be equal to the original resultant. 

Let the pairs be o'b and o"b, o"c and o'"c, and o'a and o'"d. 
The forces o'b and o"b are equal and directly opposed, hence 
their resultant is zero; for the same reason the resultant of o"c 
and o'"c is zero; hence the resultant of o'a and o'"d must be the 
resultant of the system. The resultant of the system, AD, there¬ 
fore acts through the point o Iy , the point of intersection of o'a 


230 


CIVIL ENGINEERING. 


and o'"d , and is parallel to A'D'. This is evident from the force 
polygon, since A'O and OD' are the components of A'D' in that 
polygon. The lines o'a and o'"d are in this case called the 
closing lines of the polygon o'o"o'"o lx . 

The polygon o'o"o'"o lx is called the equilibrium polygon. 

From the construction it follows that the point of the result¬ 
ant of any system of non-concurrent forces not in equilibrium 
may be determined by finding the point of intersection of the 
two closing lines of the equilibrium polygon. 

Problem IV .—To determine the intensities oj two unknown 
parallel jorces in a system oj non-concurrent forces in equilibrium 
when only the points oj application oj the unknown jorces are given. 

In Fig. 52 let the forces AB, BC, and CD whose intensities 
are given by the force polygon A'B'C'D' form a non-concurrent 
system of forces in equilibrium with two unknown parallel forces 



6 , 

1 

E 

T 

C 

Fig. 52. 

which act through the points a and b. If the unknown forces 
are parallel, they must also be parallel to the resultant of the 
forces AB, BC, and CD in order that the sum of the horizontal 
and the sum of the vertical components of the system shall each 
be equal to zero. 

In the force polgyon draw the line A'D'; it will be equal 
and parallel to the resultant of the forces AB, BC, and CD, 
and must be equal to the sum of the intensities of the two un- 









DETERMINING THE STRESSES IN A TRUSS. 


23 1 


known forces. By hypothesis the lines DE and EA drawn 
through the points a and b parallel to A'D' must be the action 
lines of the unknown forces. 

When the position of the resultant of the known forces can 
be determined without constructing an equilibrium polygon we 
may employ the following method to determine the intensities 
of the unknown forces: 

Let AB (Fig. 53) be the action line of the resultant of the 
known forces, and BC and AC be the action lines of the two 
unknown parallel forces. Let A'B' be the intensity of the force 
AB. From any point on AB, as e, draw dej perpendicular to 



AB. Then will ej and de be the lever-arms of the forces AC 
and BC with respect to any point on AB. Since for equilibrium 
the resultant of BC and CA must be equal and opposed to AB, 
we have, from the principle of parallel forces, 

Intensity BCXed = intensity CA Xe). . . (453) 

♦ 

To determine the unknown intensities of BC and CA, from 
B' draw j'd' making any convenient angle with A'B'. Lay off 
j'e' =} e , and e'd' = ed, and draw d'A' and parallel to it e'C'. Then 
will B'C' be the intensity of the force BC, and C'A' the intensity 
of the force CA; since from the similar triangles we have 

B'C':C'A' = e'j':e'd'::ef:ed, 

or B'C'Xed:\A'C'Xej .( 454 ) 

When the position of the resultant of the known forces can 
be determined only by constructing an equilibrium polygon we 
must employ the following method: 









232 


CIVIL ENGINEERING. 


In Fig. 52 assume O as the pole and draw the lines OA', 
OB', OC', and OD'. Then from any point on the action line 
of the force AB, as o', construct the equilibrium polygon 
o'o"o'"o lw o w . 

Since the action line of the force EA, acting through a, inter¬ 
sects the line o'o y at o y , we may lay off from the point o v , on the 
line o y o', an intensity equal to OA' and consider this as one of 
the components of the unknown force EA. In the same manner 
we may lay off from o iy on o Iy o'" an intensity equal to D'O, 
and consider it one of the components of the unknown force DE 
acting through b. 

In the concurrent system of forces acting at o y one of the 
angles of the equilibrium polygon 0'0"0"'o iy o y , we have the 
component of EA acting along o y o' fully given, and the action 
line of the component acting along o y o iy , and the action line of 
the force itself, EA ; hence we may determine the intensities of 
the unknown component and the force EA by drawing lines 
through the points O and A' of the force polygon, parallel to 
o y o iy and EA. In a similar manner we may determine the in¬ 
tensity of the force DE and of its component acting at o iy along 
o iy o y by drawing lines in the force polygon through O and D ' 
parallel to o iy o y and ED. We thus determine the point E' and 
the intensities of the two unknown forces; E'A' is the intensity 
of the force acting through a, and D'E' is the intensity of the 
force acting through b. In this case the line o iy o y is the closing 
line of the equilibrium polygpn. 

Hence we see that to find the intensities of two unknown 
parallel forces of a non-concurrent system in equilibrium we 
need only draw, in the force polygon, a line through the pole 
parallel to the closing line of the equilibrium polygon. It will 
divide the resultant into two parts, each of which represents 
the intensity of one of the unknown forces. 

Problem V.— To determine the intensities oj two unknown 
forces in a system oj non-concurrent forces in equilibrium when 

1 

only the points oj application of both unknown forces and the 
action line of one of them are given. 

Let the full lines AB, BC, and CD (Fig. 54) be the action 
lines of the known forces whose intensities are given by the 
force polygon A'B'C'D'\ let the full line DE be the action 



DETERMINING THE STRESSES IN A TRUSS. 


2 33 


line of one of the unknown forces, and let o' be a point on the 
action line of the other unknown force. 


E 



I. When the action line of the resultant of the known forces 
can be determined without the aid of the equilibrium polygon 
we may use the following methods: 

ist. When the action line of the resultant and that of one 
of the unknown forces intersect within the limits of the draw¬ 
ing. In this case the action line of the other force may be de¬ 
termined by connecting its given point of application with the 
point of intersection. Thus, in Fig. 54, we construct AD parallel 
to A'D', the resultant of the known forces, and find its point of 
intersection with DE; the action line of AE will pass through 
0' and this point of intersection. This construction results from 
the fact that three forces in equilibrium must form either a parallel 
or a concurrent system of forces. Having given the action lines 
of the unknown forces, their intensities may be found from the 
force polygon. 

2d. When the action line of the resultant and that of one 
of the unknown forces do not intersect on the drawing. Since 
the moment of the resultant with respect to 0' must be equal to 










234 


CIVIL ENGINEERING. 


moment of DE with respect to the same point, we may employ 
a method similar to that shown in Fig. 53. 

In diagram IV, Fig. 54, let 0' represent the point 0' of dia¬ 
gram II, AD represent the action line of the resultant AD, and 
DE the action line of the unknown force DE\ then will o'a be 
the lever-arm of the resultant, and o'b be the lever-arm of the 
force DE with respect to the point o'. In diagram III, Fig. 54, 
make A'D' equal A'D' in diagram I, and make o'b and o'a 
equal to o'b and o'a in diagram IV. Connect o' with A' and 
draw parallel to it the line aE' ; then will E'A' be the intensity 
of the force DE. Since from the similar triangles o' A'D' and 
aE'D' we have 


A'D':A'E' 


:o'D':o'a: \o'b:o'a , 


hence A'D' Xo'a = A'E' Xo'b .( 455 ) 

Having determined the intensity DE, in I lay off D'E par¬ 
allel to DE and equal to E'A' (III); A'E' will be parallel to the 
action line of EA and will represent its intensity. 

II. When the action line of the resultant cannot be deter¬ 
mined without the use of the equilibrium polygon we must 
employ the following method: 

Assume the pole O and draw the lines OA', OB', OC , and 
OD'\ then from the preceding problem OA' may be taken as 
one of the components of the force EA whose direction and 
intensity are required, and D'O as one of the components of the 
force DE. From o', the known point on the action line of the 
force EA, as an origin, construct the equilibrium polygon 
o'o"o'"o lv o v . The closing line oV will be the action line of the 
other component of the forces DE and EA. Hence if in the 
force polygon we draw OE' parallel to o v o', and D'E' parallel to 
DE, E'A' will be the intensity and direction of the force EA, 
and D'E' will be the intensity of the force DE. To complete the 
solution the line EA should be drawn through o' parallel to 
E'A'. 

It will be observed that when only one point on the action 
line of an unknown force is given, the equilibrium polygon must 
begin at that point. 


DETERMINING THE STRESSES IN A TRUSS. 


2 35 


Uses of the Equilibrium Polygon. —In a system of non¬ 
concurrent forces, to determine whether the forces are in equi¬ 
librium or have a resultant, we must utilize both the force and 
equilibrium polygons. If the force polygon closes, the sum of the 
horizontal and the sum of the vertical components of the system 
must each be equal to zero. If the equilibrium polygon closes, 
the sum of the moments of the forces with respect to any point 
in the plane of the forces must be equal to zero. Hence if the 
force and equilibrium polygons of a non-concurrent system of 
forces both close, the system is in equilibrium, and conversely 
the force and equilibrium polygons of a non-concurrent system 
of forces in equilibrium must close. 

If the resultant of a system of forces is a single couple, the 
force polygon will close, but the equilibrium polygon will not. 
Hence we may assume that a non-concurrent system whose force 
polygon closes but whose equilibrium polygon does not may 
be reduced to a single couple. 

The Solution of a Truss. —Let it be required to determine 
by the graphic method the maximum and minimum stresses 
in the truss shown in Fig. 55. The truss is fixed at the right- 



hand support, rests on a roller at the left support, and is sub¬ 
jected to a vertical load of 500 pounds at the end panel points 
and a vertical load of 1000 pounds at each of the other panel 















236 


CIVIL ENGINEERING. 


points of the upper chord, and to an intermittent wind pressure 
equivalent to a normal load of 500 pounds at the ridge and eave 
panel points, and of 1000 pounds at the intermediate points. 

Case I.— When the truss is acted upon by vertical weights alone. 

In Fig. 55 the reactions are each equal to 3000 pounds, and 
act vertically. The position diagram is lettered according to 
the method employed for concurrent forces known as Bow’s 
notation, and the force polygons are constructed for each panel 
point by taking the forces and stresses in order clockwise, as 
indicated by the arrow. 

Force Polygon of External Forces.—Beginning at A’ con¬ 
struct the force polygon of the external forces, A'B'C'D'E'F ' 
G'H'I'A'; in this polygon the distance C'D f represents 1000 
pounds. 

Panel Point ABCJ.—At this panel point the forces AB and 
BC are known, and the action lines of the stresses CJ and JA ; 
the intensities and directions of the stresses CJ and J A are un¬ 
known. Construct the force polygon A'B'C'J'A' . Then C'J' 
is the direction and intensity of the stress CJ, and J'A' is the in¬ 
tensity and direction of the stress JA. As CJ acts towards the 
panel point it is compressive, and as JA acts away from the 
panel point it is tensile. 

Panel Point CDKJ.—At this point we know the compressive 
stress JC and the force CD; the intensity and direction of the 
stresses DK and KJ are unknown. Since the stress JC is com¬ 
pressive, it must act towards the panel point; hence we construct 
the force polygon J'C'D'K'J' . Then D'K ' and K'J' are the 
intensities and directions of the stresses in DK and KJ. As 
they both act towards the panel point they are both compressive 
stresses. 

Panel Point JKLA.—At this panel point the tensile stress 
AJ and the compressive stress JK are known. Hence we con¬ 
struct the force polygon A'J'K'L'A' and find the intensity and 
direction of the stresses KL and LA. As both act away from 

r 

the panel point they are both tensile. 

Panel Point DEMLK.—At this panel point we know the 
tensile stress LK , the compressive stress DK, and the force DE; 
hence we construct the force polygon L'K'D'E'M'L' and find 
the intensity and direction of the stresses EM and ML. As 



DETERMINING THE STRESSES IN A TRUSS. 


237 


both act towards the panel point, they are both compres¬ 
sive. 

Panel Point MEFN.—At this point we know the compres¬ 
sive stress ME and the force EF; hence we construct the force 
polygon M'E'F'N'M', and find the intensity and direction of 
the stresses FN and NM. The stress FN acts towards the 
panel point and is compressive, the stress NM acts away from 
it and is tensile. 



Panel Point ALMNO.—At this point we know the tensile 
stresses AL and MN and the compressive stress LM ; hence we 
construct the force polygon A'L'M'N'O'A' and find the com¬ 
pressive stress NO and the tensile stress OA. 

Panel Point ONFGP. —At this point we know the compres¬ 
sive stresses ON and NF and the force FG; hence we construct 
the force polygon O'N'F'G'P'O' and find the compressive stress 
GP and the tensile stress PO. 

Panel Point AOPQ. — At this point we know the tensile 
stresses AO and PO; hence we construct the force polygon 
A'O'P'Q'A' and find the compressive stress PQ and the tensile 
stress QA. 

Panel Point QPGH. — At this point we know the compres¬ 
sive stresses QP and PG and the force GH ; hence we construct 
the force polygon Q'P'G'H'Q' and find the compressive stress 













238 


CIVIL ENGINEERING. 


HQ. If the solution is correct, the line H'Q' must connect 
the points Q' and H', and be parallel to HQ. 

In the position diagram, compressive stresses should be in¬ 
dicated by heavy lines as soon as the character of the stress in 



a member is determined. The accuracy of the results will in¬ 
crease with the scale of the drawing and with the skill of the 
draughtsman. 

Case II .—When the truss is subjected to vertical loads and 
to wind pressure acting on the lejt side. 

Beginning at the force BC (Fig. 56), construct the lines of the 
force polygon of the external known forces, B'C'D'E'F'G'H' /' J' 
K'L'M'. As the truss rests on a roller at the left end, we know 
that a vertical line through B' will represent the direction of 























DETERMINING THE STRESSES IN A TRUSS. 239 

the reaction at the panel point ABCDN, and that the point A' 
lies somewhere on this line. To determine A', we might make 
use of Problem V. A simpler method, however, is the following, 
in which the reactions at the panel point due to the vertical and 
oblique loads are considered separately. From inspection it is 
evident that the resultant of the wind load is perpendicular to and 
bisects the member FO. This resultant with the reactions at the 
supports due to the wind pressures alone must form a concur¬ 
rent system of forces, since the resultant of the reactions must 
be equal and directly opposed to the resultant of the forces pro¬ 
ducing them. Since, by hypothesis, the reaction at the panel 
point ABCDN is vertical, the point V must be the point where 
the action lines of the resultant and the reactions due to wind 
pressure intersect. If we lay off the distance VA " by the scale 
equal to 3000 pounds, the total wind pressure, A"W and WV 
must represent graphically the intensities of the reactions due 
to wind pressure at the left and right supports respectively. If 
we lay off to scale WB" equal to 3000 pounds, the reaction 
at the panel point ABCDN due to the weights alone, then 
A"B" must be the total reaction at the panel point ABCDN. 
Making B'A' of the force polygon equal to B"A ", we determine 
the point A' and the reaction M'A'. 

We then determine the intensity and the character of the 
stress in each piece, by constructing in succession the force poly¬ 
gons A'B'C'D'N'A ', N'D'E'F'O'N ', A'N'O'P'A', P'O'F'G'H' 
Q'P’, Q'H'I'J'R'Q ', A'P'Q'R'S'A', S'R'J'K'T'S', A'S'T'U'A 

and U'T'K'L'U'. 

Case III.— When the truss is subjected to vertical loads and 
to wind pressure acting on the right side, as shown in Fig. 57* 

As before construct the lines of the force polygon of the known 
external forces B'C'D'E'F'G'H'FJ'K'L'M the point A' must 
be on the vertical line through B '. 

The reactions may be determined by constructing an equi¬ 
librium polygon as in Problem V, with the panel point AUKLM 
as the initial vertex. However, it will be simpler to find the 
vertical reaction through the panel point ABCN in the follow¬ 
ing manner: From F" draw the force polygon F"H"J''L"M' f 
oAhe wind pressures FG, HI, JR, and LM ; the reaction at 
the panel point ABCN due to the wind pressure will be the 


240 


CIVIL ENGINEERING. 


vertical line A"F ", whose length must be determined. Assume 
O as a pole and draw the lines OF ", OH", OJ", OL", and 
OM". Prolong the lines of wind pressure, and beginning at o l 
construct the equilibrium polygon o 1 o 2 o 3 o 4 o 5 . Draw the closing 
line o 5 o\ and parallel to it the line OA" of the force polygon. 
Then A"F" will be the vertical reaction at the panel point 



B 


11 n 1 11 111 r~ 
1000 0 


-I-1 

10C00 lbs. 


5u00 


Fig. 57. 


ABCN due to the wind pressure. The vertical reaction at this 
panel point due to the weight alone is 3000 pounds. Laying 
off F"B" equal to 3000 pounds, the total reaction at the panel 
point ABCN is A"B". Making B'A' equal to A"B" and draw¬ 
ing the line M'A', the force polygon of the external forces is fully 
given. 

The intensity and character of the stresses may be determined 
by constructing in succession the force polygons A'B'C'N'A', 
























DETERMINING THE STRESSES IN A TRUSS. 


241 


N'C'D'O'N ', A'N'O'P'A', P'O'D'E'Q'PQ'E'F'G'R'Q', 
A'P'Q'R'S'A', S'R'G'H'PT'S', A'S'T'U'A', and U'T'PJ'K'U'. 

The three force polygons, Figs. 55, 56, and 57, give us the 
greatest and least stresses in the different members of the truss 
under the varying conditions of the loading, and hence all the 
data necessary for the designing of the members. 

In solving a truss by the graphical method the following points 
should be noted: 


1 st. The external forces should be indicated as acting in 
their proper directions and at their proper points of application, 
outside the perimeter of the truss. 

2d. In constructing force polygons for the different panel 
points, always take the forces and stresses in the same order 
clockwise or counter-clockwise, beginning with the known forces 
and stresses. 

3d. In a graphic solution all redundant or unnecessary mem¬ 
bers of a truss should be omitted. 

4th. If the truss is fixed at both ends, the reactions are both 
parallel to the resultant of the known external forces and can 
be determined by the methods explained under Problem IV. 


CHAPTER XIII. 


THE GRAPHIC METHOD APPLIED TO PARALLEL 

FORCES. 

Besides the solution of simple trusses without parallel 
chords, there are several other engineering problems which can 
be solved more easily by the graphic than by the analytic 
method. 

I. To find the center of gravity oj an area. According to the 
theory of flexure the neutral axis of the cross-section of a beam 
passes through its center of gravity, and the stress in the extreme 
fibers of the cross-section depends upon the distance of these 
fibers from that axis. To determine the strength of a beam to 
resist flexure we must therefore be able to determine the posi¬ 
tion of the neutral axis or the center of gravity. 

7 / the cross-section is symmetrical with respect to two rect¬ 
angular axes , as a rectangle, circle, ellipse, an I section, 
etc., the center of gravity must lie at the intersection of these 
axes. 

7 / the cross-section is symmetrical with respect to only one 
axis, as a T section or an angle with equal legs, we know that 
its center of gravity must lie on that axis, but its actual position 
is not known. It may be determined graphically in the following 
manner, providing the cross-section can be divided into smaller 
areas, the positions of whose centers of gravity are known, and 
whose areas may be easily computed. 

Let it be required to find the center of gravity of a T beam 
whose flange is 3j"Xf" and whose web is 3"Xf". The cross- 
section is shown in Fig. 58, is symmetrical with respect to AB r 
and may be divided into two rectangles whose areas are and 

square inches respectively. 

Through the center of gravity of the first rectangle draw the 
action line of its resultant force of gravity Coi ; through the center 

242 




THE GRAPHIC METHOD APPLIED TO PARALLEL FORCES. 243 




Y 


4 


c' 






of gravity of the second rectangle draw the action line of its 
resultant force of gravity EF. Then will the point where the 
resultant of Co\ and EF intersects 
AB be the center of gravity of the 
cross-section. 

Construct the force polygon 
C'D'E'F ', in which C'D' is 35 
units and E'F ' 30 units in length. 

Assume the pole O, and draw the 
forces OC', OD', and OF'. From 
any point on the action line of 

Co 1, as 01, construct the equilibrium polygon O1O2O3. The 


! ] j~3~ *1 


t 

y%- 




' 0 <; 


TOo 


l F " 


D'E 


N 


F' 


Fig. 58. 


resultant of Co\ and FF will pass through 03, be parallel to 
C01 and EF, and intersect A B at the center of gravity of the 
area. 

If the cross-section is not symmetrical with respect to any axis, 
as an angle with unequal legs, the above operation may be 
repeated. We may first find the resultant force of gravity per¬ 
pendicular to the axis of one of the legs, and then find the 
resultant force of gravity perpendicular to the other; the two 
resultants will intersect at the center of gravity of the cross- 
section. 

The center of gravity of any irregular area used in the design 
of beams may be found by dividing the total area into rectangles, 
triangles, or circular sections. The center of gravity of any two 
areas must lie on the line joining their centers of gravity. The 
center of gravity of a triangle lies on the line connecting any 
vertex with the middle point of the opposite side, and two-thirds 
of the distance from the vertex. The center of gravity of a cir¬ 
cular sector lies on the bisecting radius and at a distance from 

f (radius X chord) 

the center = —--. 

arc 

II. To find the bending moment at any section 0} a beam. In 
the force polygon of a system of parallel forces the perpendicular 
distance from the pole to the line representing the original forces 
is called the polar force. Thus in Fig. 58 the perpendicular 
distance from O to C'F' is the polar force. The graphical de¬ 
termination of the bending moment depends on the following- 
theorem : 












244 


CIVIL ENGINEERING. 


The moment oj any force with respect to any point in its plane 
is equal to the polar force multiplied by the distance between the 
components of the force in the equilibrium polygon measured on 
a line through the point parallel to the action line oj the force. 

Let AB, Fig. 59, be the action line of any assumed force, 
and a be the point about which the moment is taken. Let A'B' 

be the intensity of the force, OA' and OB' 
its two components, and OH the polar force. 
From a draw ab perpendicular to AB; from 
b draw be parallel to OB', and bd parallel 
to OA'; through a draw cd parallel to AB. 
Fig. 59> Then, according to the theorem, 


c 

•“v 

I ' 

a!-5* 

I / 

1 x 

1 / 

V 

d 


b o^-— 


rB 


"1A 


H 


A'B' Xab = OHXcd. 


From the similar triangles OA'B' and bed we have 

A'B' :OIT : :cd:ab or A'B' Xab = OHXcd, . (456) 

which was to be proved. 

Let it be required to find the bending moment at any section 
of the beam shown in Fig. 60 resting on end supports and acted 



R i 

Fig. 60. 


upon by the parallel forces ED, DC, and CB, whose intensities 
are 5, 15, and 10 pounds. The reactions will be vertical. 

Construct the force polygon of the known vertical forces 
E'D'C'B'. Assume a pole O and draw the forces OB', OC ', 
OD', and OE'. From any point on the action line of either 



















THE GRAPHIC METHOD APPLIED TO PARALLEL FORCES. 245 


reaction, as 0 1, construct the equilibrium polygon 0102030405. 
The closing line o 5 o\ will determine the direction of OA' of the 
force polygon, and will determine the intensities of the reactions 
B'A' and A'E'. The bending moment at b is 

M = B'A'Xab, . (457) 

but by the theorem above 

B'A' Xab = HOXo 2 j .(458) 

The bending moment at c is 

M"=B'A'Xac-C'B'Xbc , .... (459) 


but according to the theorem above 

B'A'Xac = HOxig and C'B'Xbc = HOxio 3 . . (460) 

Hence 

B'A'Xac-C'B'Xbc = HOxig-HOxio 3 = HOXosg. . (461) 

The bending moment at d is * 

M'" = B'A'Xad-C'B'Xbd-D'C'Xcd. . . (462) 

But B'A'Xad = HOxlh ..(463) 


C'B'Xbd = HOXlk, . (464) 

D'C' Xcd = HO X ko^ .(465) 

Ui—(I k T ko 4) = 0 4.h, • • • • • (466) 

Hence 


B'A ' X ad - C'B' Xbd- D'C' Xcd=HOX 0J1. . (467) 

Hence at /, g, and h the bending moment is equal to the 
polar force multiplied by the distance between the opposite sides 
of the equilibrium polygon measured on the action lines of the 
forces. In a similar manner, it could be shown that at any 






246 


CIVIL ENGINEERING. 


other section of the beam the bending moment will be equal to 
the polar force multiplied by the distance between the opposite 
sides of the equilibrium polygon, measured on a line drawn 
through the section considered and parallel to the action lines 
of the parallel forces. We may therefore always determine the 
bending moment at any section of a beam by the graphic method 
when all the normal forces acting upon it can be determined. 

III. To determine the bending moment on a pin. If we know 
the stresses acting in the members of a truss which are connected 

e by a pin, as Fig. 61 a, we can resolve them 

20,000 v ..F J 1 > o 

/ 42,426 into horizontal and vertical components and 
// determine the bending moment at any cross- 

D 1>o,ooo a ^00 B section due to either system of components as 

described above. AC is at the middle point 
' r io,ooo °f the pin; AF is made of two bars each ij 

~ , inches from AC: AE is made of two struts 

Fig. 61 a. t 7 

each 2\ inches from AC; AD is made of two 
bars each 4J inches from AC; AB is made of two bars each 5J 
inches from AC. The angle EAF is 45 degrees. 

From the principles of Mechanics we know that if at any 
section the horizontal moment be represented by M; t and the 
vertical moment by M v , the resultant moment will be y/Mjf + Mf. 
If, in constructing the force and equilibrium polygons of the hori¬ 
zontal and vertical components, we employ the same polar force, 
we shall have for a cross-section at a distance x from the end 
of the pin 

Mu — polar force Xh, 

in which h = distance between sides of the equilibrium polygon 
of the horizontal components at x; 

Mv — polar force Xv, 

in which v = distance between sides of the equilibrium polygon 
of the vertical components at 

Hence VMj? + M V 2 = polar force V 7 t 2 + z/ 2 . 

The maximum resulting moment will be at the section where 
V h 2 + v 2 is a maximum. 









THE GRAPH ^ METHOD APPLIED TO PARALLEL FORCES. 247 


Let these principles be applied to the determination of the 
maximum bending moment in the pin represented in Fig. 61. 

The total length of the pin between the exterior eye-bars 
is 10J inches, as shown in I, Fig. 61. The distances between 
the axes of the different pieces is shown in I and III, Fig. 61. 
The force polygon of the horizontal components is II, Fig. 61, 
and the equilibrium polygon is I, Fig. 61. 


30,000 

A 


*%r- 


HORIZONTAL COMPONENTS 


■\\ 

\ 

\ 

1 \ 


B 

-3”- 


15.000 

A 


30,000 

A 


15.000 

A 


-3-- 


—3-- ><%’> 


^2 

\ 

\ 


a 


45,000 




a' 


\'k 5 

HV-V^o 
'\- 


/0 

^ 2 , 
' 1 I 

1 o 1 
1 

C'd' 


III 


4 /^ 

/ 




A 

a / 

/ 

/5 


Y 


/ 


K l>f 

Q' u 


-i Yi-Y±- 
1 ]c 3 ' 

XV !.L^"2 

1 v ri5|ooo 

1 I 


G 


3 JT' 

! 4 

4 » 


45,000 


/ 

5 / 

II U/ 

i / 3 

o *-4 
\\ 6 


I k j a I 1 


10,000 10,000 10,000 
VERTICAL COMPONENTS 

Fig. 61. 


\ 2 IB' 

i\ 


iA 


d' 


c'f' 


The force polygon of the vertical components is IV, Fig. 61, 
and the equilibrium polygon is III, Fig. 61. 

To obtain symmetrical equilibrium polygons, the pole of the 
force polygon must be opposite the middle points of the lines 
E'A' and I'G'. 

Then at any point of the pin, as the point of application of 
one of the bars AF, the horizontal moment will be OC'Xab , 
and the vertical moment will be OH'Xcd ; or, since OC'= 
OH' by construction, the resultant bending moment at this point 
of the pin will be OC' V( ab ) 2 + (cd) 2 . 

This may be represented graphically by drawing a'b' in 
V, Fig. 61, equal to ab, perpendicular to it c'd' equal to cd, and 
oc' equal to OC' in diagram II. The area ooa'c' will be the re¬ 
sulting bending moment in the pin at the action line of the force 
















248 


CIVIL ENGINEERING. 


BC. In the same manner we can determine the bending moment 
at any other point. 

IV. To find the moment of inertia 0} an area approximately. 
Let it be required to find the moment of inertia of the angle 
represented in Fig. 62 about a vertical axis through its center of 
gravity G. 



Fig. 62. 


Let the area be subdivided into three rectangles whose 
centers of gravity are on the vertical lines through A, B, and C. 
In the force polgyon A'B'C'D' let A'B' represent the intensity 
of the gravity force acting on the line through A, B'C' the inten¬ 
sity of force acting at B, and C'D' the intensity of force acting 
at C. Let / be the moment of inertia of the angle about the 
axis GE. Then we shall have 


I=A'B'X (AG ) 2 + B'C' X (. BG ) 2 + C'D ' X (CG) 2 , approximately. 

Assume the pole O, and draw the rays OA', OB', OC', and 
OD'. Construct the equilibrium polygon whose sides are 1, 2, 3, 
and 4. Then from the preceding theorem we have 

A'B'xAG = EFxOH, .*. A'B'X (AG ) 2 = EFxAGxOH ; (468) 

B'C'XBG = FK X OH, B'C'X(BG ) 2 = FKxBGxOH; (469) 

C'D'XCG = EKXOH, .*. C'D'X(CG ) 2 = EKxCGxOH. (470) 
But 

EFxAG = twice the area of the triangle OiEF, 

FKXBG= “ “ “ “ “ “ o 2 FK, 

FK X CG = “ “ “ “ “ “ o 3 EK. 


Hence the moment of inertia of the cross-section 











THE GRAPHIC METHOD APPLIED TO PARALLEL FORCES. 249 

= [EFXAGX 0 H] + [FKXBGX 0 H] + [EKXCGX 0 H] 

= twice the area of the equilibrium polygon X OH. 

The error introduced in this method is that of assuming 
that the radius of gyration of each rectangle about GE is equal 
to the distance of its center of gravity from the axis, or R = d, 
when in fact R = \ / d 2 -\-r 2 . For the area whose center of gravity 
is on Ao \, 

V 

R = radius of gyration of area about GE , 
r— radius of gyration of area about Aoi , 
d = distance AG. 

In theory the line 01O2O3O4 is a curve which is obtained 
by making the width of each of the elementary parts into which 
the section is divided equal to dl , an elementary part of the length 
of the cross-section. It is evident that we will approach this 
limit as we increase the number of parts into which we divide 
the cross-section; we thus decrease the radius of gyration r. 

This affords a method of determining very closely the mo¬ 
ment of inertia of irregular sections. By drawing the cross- 
section to a large scale, dividing it into many parts, usually of 
equal width, and measuring the area of the equilibrium poly¬ 
gon with a planimeter, the moment of inertia may be deter¬ 
mined. 

V. To construct an equilibrium polygon which shall pass 
through two or three given points. Having given the three par¬ 
allel forces BC, CD , and DE in Fig. 63, let it be required to con¬ 
struct an equilibrium polygon passing through the points a and 
b. The assumed forces being parallel, by means of the force 
polygon A'B'C'D'E'A' and the equilibrium polygon 12345 we 
may determine the intensities of the parallel forces A'B' and 
E'A' which, acting at the points a and b, will hold the given forces 
in equilibrium. If through A' the line A'X' is drawn parallel 
to the line connecting a and b, and a pole is taken anywhere on 
A'X' for a new equilibrium polygon beginning at a, this new 
polygon will have vertices at a and b, since its closing line, 5, 
must pass through a, the origin of the equilibrium polygon, and 
be parallel to A'X', which by construction is parallel to a line 
connecting a and b. 



250 


CIVIL ENGINEERING. 


To construct an equilibrium polygon passing through three 
points a, b, and c not in a straight line, the force polygon for all 
the known forces is constructed. Selecting a pole and construct¬ 
ing an equilibrium polygon, the line A'X corresponding to the 
points a and b may be determined. Selecting a second pole and 


L i2)_-r _ 

. At Cfr-P 




a — 




T‘ 

I 


- 




i i 

's'c * 


T ^ 
y 

/ 


^ -X 




/ i 

^ i 


!/ 


Fig. 63. 


constructing an equilibrium polygon the line A ,r X for the points 
b and c may be constructed; the point where the lines A'X and 
A"X intersect will be the pole, by the use of which the desired 
equilibrium polygon may be constructed. 

The methods above explained also apply to a system of non¬ 
parallel forces. 








CHAPTER XIV. 


SOLUTION OF TRUSvSES CONTINUED. 



Modified Pratt Truss.—In the modified form of this truss, 
Fig. 64, the end vertical is omitted, and the end diagonal becomes 
the batter-brace. This modification makes 
the last vertical a tie, which supports only 
the load suspended from its lower end. 

The Baltimore truss (Fig. 65) is a Pratt 
truss of great height and span which has 
been modified by increasing the number of panel points of the 
lower chord by inserting half-verticals between the main ones. 
If the upper chord is not straight, but of the bowstring type, 
it is a Petit truss. 


Fig. 64. 



The Cantilever Truss.—The simple cantilever truss (Fig. 66) 
is a truss which has a single support at one of its ends, to 

which its chords are fastened. A 
double cantilever rests on a single 
central support; as the two over¬ 
hanging parts balance each other, 
the chords are not fastened to the 
support. The cantilever truss may 
be solved by the analytic or graphic 
methods in the same manner as the truss resting on two sup¬ 
ports. The solution must, however, be begun at the free end 
of the cantilever. If the cantilever supports only weights, the 
stress in the upper chord will be tension, and in the lower chord 
compression. 



251 
































252 


CIVIL ENGINEERING. 


Combined Cantilever and Simple Truss.—A truss resting on 
two supports may have one or both of its ends prolonged so as 
to form cantilevers, Fig. 67. In solving this truss, the first step 


4 8 4 8 



is to determine the reactions at the supports by either the method 
of moments or by the graphic method. When the reactions are 
known the solution of the truss is begun at one of the free ends. 
If a truss of this description supports only weights, the chord 
stresses in the members between the supports may be either 
compressive or tensile, depending on the system of loading. 

Continuous Truss.—A continuous truss, Fig. 68, is one which 
rests upon three or more supports; it is usually a riveted truss. 
In determining the reactions at the supports, if uniformly loaded, 
such a truss may be treated as a beam resting on three or more 



Fig. 68 . 

supports. Thus if it rests on three supports equally spaced, and 
its load and dead weight are uniformly distributed, the reaction 
at each end support will be &, and the reaction at the middle 
support will be hb of the total weight and load. If uniformly 
loaded it may be treated as made up of separate trusses resting 
on end supports. 

Bowstring Truss. — A bowstring truss (Fig. 69) is one in 
which one of the chords is straight and the other is a circle or 
parabola or made up of the chords of these curves. A double 
bowstring truss is one in which both chords are thus bent and 
are concave to each other. 

In a bowstring girder the shear is divided between the 














SOLUTION OF TRUSSES. 


253 


inclined chord and the web members; the total stress in the 
former is therefore increased, and in the latter decreased. In 
a uniformly loaded truss the curvature of the chord may be so 
adjusted as to reduce the stress in the inclined web members to 
zero. If there is no stress in the inclined web members, the 



stress in each of the vertical web members must be the same, 
and the stress in all the members of the straight chord must also 
be the same. 

If in Fig. 69 we assume that there is no stress in the diagonal 
web members, equilibrium will result if the equations of mo¬ 
ments below are satisfied. 

Let / = length of truss AF] 

V = panel length AB 
h = height of truss EE" ; 

JT = DD "; 
h" = CC "; 
h"' = BB". 

Cutting the lower chord between D" and E" and taking moments 
about E of the upper chord, 

7 W X 4/' - 2W X 3/' -2WX2V- 2 W X /' 

+ stress D"E" (or AB") Xh = o; 

WV 

therefore /,= -l6 stress AW’ . (47l) 

Cutting the lower chord between C" and D" and taking moments 
about D of the upper chord, 

7 W X3/' -2 W X 2V -2 WV + stress C"D "(or AB")Xh' = o; 

wr 

^ ~ 1 ^stress AB" 


therefore 


( 47 2 ) 
















254 


CIVIL ENGINEERING. 


Cutting the lower chord between B" and C" and taking moments 
about C of the upper chord, 

jW X 2/' - 2W X r + stress B"C" (or AB") X h" = o; 


therefore 


h" 


WV 

1 "stress AB' r * 


( 473 ) 


Cutting the lower chord between A and B " and taking moments 
about B of the upper chord, 


therefore 


yW/' + stress AB" Xh'" = 0; 
WV 


h m = - 7 


stress TiF 


(474) 


Hence to fulfil the conditions that there shall be no stress in 
the diagonal web members, the relative values of h, JV, h", and 
k in the truss shown in Fig. 69, must be as 16, 15, 12, and 7. 

To support a moving load such a truss would require diag¬ 
onal main braces and counterbraces. The bowstring girder is 
more easily solved by the graphic than by the analytic method. 

The Pegram truss (Fig. 70), which is a modification of the 
bowstring truss, is formed by drawing the arc of a circle through 



the points A and B of the batter-braces, and making the inter¬ 
mediate upper chord members equal to each other and equal in 
number to those of the lower chord. 

Compound Truss.—A compound truss (Fig. 71) is one which 



has two or more systems of web-bracing. In computing the 
stresses in the members of such a truss, resting on end supports, 
it is to be assumed that the loads at B and D are carried by 










SOLUTION OF TRUSSES. 


255 


a truss composed of the full-lined web members with the chords 
and end-posts, and that the loads at A, C , and E are carried by 
a truss composed of the dotted web members with the chords and 
end-posts. The stresses in the chord members and end-posts are 
therefore the sum of the stresses deduced for the separate trusses. 

If the web system is a riveted lattice-work, it becomes a 
lattice truss and may be treated as a plate girder. 

Three-hinged Truss.—A three-hinged truss (Fig. 72) is a 
truss formed of two separate trusses which rotate about fixed 
axes at the supports, and abut against a common pin at their 
free ends. This construction allows the truss to rotate about the 


E f 



axes and thus accommodate itself to changes of length of its com¬ 
ponent parts clue to variations in temperature. The fixed axes 
are connected by a tie-rod, or are fixed by strong abutments. 

If a concentrated load acts on one side of the truss only, as 
at D , the reaction at C must pass through A if we neglect the 
friction of the joints and the weight of the truss. The action 
lines of the reactions will therefore be BE and CE. The action 
lines of the reactions will have the same directions if ED is the 
action line of the resultant of uniformly or non-uniformly distrib¬ 
uted load resting on the left half of the truss. Knowing the 
intensity of the force acting at D , we may readily determine the 
intensities of the reactions. 

Similarly if the load is placed on the right half only, as at G , 
we may find the reactions whose action lines are BF and CF. 

If both loads act simultaneously, the reaction at A will be 
the resultant of reactions acting along FA and CA, or along BA 
and EA. Having thus determined the direction of the reaction 








256 


CIVIL ENGINEERING . 


at A due to both loads, the action lines of the reaction at B 
due to both loads is obtained by prolonging the action line of 
the resultant reaction at A until it intersects ED and joining 
the point of intersection with B. 

Having determined the reactions, the stresses in the different 
members may be determined as in any other truss. If employed 
as a roof-truss, the maximum and minimum stress due to dead 
and wind loads in each member must be determined; if em¬ 
ployed as a bridge-truss, the maximum and minimum stress 
due to dead and live load in each member must be de¬ 
termined. 

The Boom Derrick.—The boom derrick (Fig. 73) is a machine 
employed in lifting weights and distributing them over a horizon¬ 



tal area whose extreme radius is the length of the boom. The 
vertical member BC is the mast; it is a wooden or steel column 
which rotates about a pin resting in a socket at C. The inclined 
member AC is the boom; it is a wooden or steel column which 
rotates about a horizontal axis at C. The member AB is the 
topping-lijt, boom-hoist , or boom-jail; it is a rope attached to the 
boom at A, passes through a pulley at B, and is fastened to a 
cleat attached to the mast at C, or passes through a pulley at the 
same point and runs thence to the hoisting-engine at J. The 
member IB is one of four or more guys which are employed to 
hold the mast in a vertical position. The weight to be lifted is 
attached to a rope called the main hoist or main jail at W; this 
rope passes through a pulley at A, through another at C, and 
runs thence to the hoisting-engine at J. To reduce the tension 
in the topping-lift rope, pulley-sheaves may be introduced at 









SOLUTION OF TRUSSES. 


257 


A and B, and to reduce that in the main fall, CJ, pulley-sheaves 
may be introduced at A and IT. 

Let angle BCA = DA W = CLK = a ; 

“ BAC=ADW=p; 

CBA =DWA = y\ 

* ; HBG = go° — y = 

W = weight whose intensity is represented by 
AW. 

Let it be required to determine the maximum stress in AC, 
AB, BC, and IB while the boom rotates from a horizontal to a 
vertical position. 

To simplify the discussion we shall first assume that the main 
fall is attached at A, and the topping-lift at B. The derrick is 
now a simple truss with one exterior force and two stresses acting 
at each vertex. Since the forces IB, AW, and the reaction at C 
are the only exterior forces acting on the derrick and are in equi¬ 
librium, they must form a concurrent system, and the reaction 
at C must pass through the intersection of IB and AW. 

The Boom.—If we resolve AW into its components in the 
direction of AC and AB, we have 


AD = W 


sin y 
sin p" 


From the triangle ABC we have 


AC\BC : :sin y.sin /?, 


AC sin x i AT , ..AC 

and AD = . . . (475) 


• * BC sin p 


BC 


sin x 

In any derrick AC and BC are fixed lengths, hence —— 7, is a 

sin rJ 

constant ratio, and the compression in the boom is constant. If 

AC = BC, AD — W. 

The Topping-lift.—The component of AW in the direction 
of AB is 








258 


CIVIL ENGINEERING. 


From the triangle ABC we have 


BA : BC :: sin a : sin /?, 


BA sin a 
BC sin /?’ 


or 


AE = W 


BA 

BC * 


(476) 


BC being fixed, the ratio 


BA 

BC 


increases as BA increases and is a 


maximum, under the condition assumed, when the boom is 
horizontal. 

The Mast.—If the stress in the topping-lift at B be resolved 
into its components, the stress in the mast will be 


_ W sin cj) sin a W cos y sin a TTT BA cos y . 

FB = -.—H- = -•—77 =lV 777;—. (477) 


sin /? 


sin p 


BC 


Since BC , the length of the mast, is constant, the stress in the 
mast due to the topping-lift will vary with AB cos y, which is the 
projection of the topping-lift on the mast. It will be compres¬ 
sive when A is below B, and will increase as the vertical distance 
between A and B increases; it will be tensile if A is above B , 
and will increase as the vertical distance between A and B in¬ 
creases. 

The Guy.—The stress in the guy due to topping-lift will 
be equal to BG : 

__ IV cos 6 sin a W sin r sin a „ T AC sin a 

BG= -r-=- A-q -( 47 8) 

sin p sinp BC K ‘ 


Since AC and BC are constant, the stress in the guy will vary 
with the sine of a, and will be a maximum when a equals 90 
degrees or the boom is horizontal. 


If the guy is not horizontal, 


W X A C sin a 
~ 1 BC~ 


will be the hori¬ 


zontal component of the stress in the guy. The stress in the 
guy will therefore be 


W 


AC sin a 
BC cos ’ 


(479) 













SOLUTION OF TRUSSES. 


2 59 


which increases as cf>' increases. The vertical component of this 
stress which must be borne by the mast is 


W 


AC sin a tan </>' 
BC 


(480) 


in which <f>' is the angle which the guy makes with the horizontal. 

If the main fall is fastened at C, the stress in the boom will 
be increased by the tensile stress in the fall, which must be W. 
If the boom-fall is fastened at C, the stress in the mast will be 
increased by a compressive force equal to the stress in the fall 
between B and C. If a single pulley is employed at B, this com- 

W sin ct 

pressive stress will be AE or - g - n ^ • If pulley-blocks are 

employed at A and B, the stress in the boom-fall between B and 
C, and the compression due to it, will be diminished. 

If the boom is fastened to the mast above C, the mast will 
also be a vertical beam resting on end supports, and subject to 
a horizontal bending force at the junction of mast and boom 
equal to the horizontal component of the stress in the boom. 
The mast will then have a shear and bending moment at every 
section. 

Reaction.—The derrick when at rest is acted on by the weight 
which is being lifted, the stress in the guy, and the stresses in 
the hoists if the hoisting-engine or other power is placed to 
one side of the derrick. The reaction at the bottom of the mast 
is a force equal and opposed to the resultant of these forces. 
If the hoisting-apparatus is attached to the mast, the derrick 
will be acted on by the weight and guy only, and the reaction 
will be opposed to their resultant. 













CHAPTER XV. 


MASONRY DAMS AND RETAINING-WALLS. 

A masonry dam is a wall designed to resist the lateral pres¬ 
sure of the water upon one of its sides. If the wall is straight, it 
must resist the water pressure by its weight alone; if curved, 
it may also resist as an arch. Only straight walls are here con¬ 
sidered. 

The side exposed to the water pressure is the back , the oppo¬ 
site side is the jace, and the bottom surface is the base. The 
intersection of the back and the base is the heel , the intersec¬ 
tion of the base and the face is the toe , and the intersection of 
the face and the top is the crest of the dam. The face and back 
may be either vertical or inclined, and if inclined, may be either 
planes or curved surfaces. The degree of inclination is called 
the batter; its measure is the tangent of the angle between the 
inclined and horizontal planes. The profile of a dam is a ver¬ 
tical section normal to the face. 

Water Pressure. — From Mechanics of Liquids we know 
that: 

I. The pressure of the water on the back of the dam is at 
all points normal to the surface. 

II. The intensity of the pressure upon any submerged plane 
surface is equal to the weight of a column of water whose base 
is the surface pressed, and whose height is the depth of the center 
of gravity of the surface below the surface of the water. 

III. The resultant pressure upon any immersed rectangular 
plane surface one of whose sides coincides with the water line, 
acts in a normal vertical plane through its center of gravity, 
and at two-thirds the depth of the lower edge below the water 
surface. 

Let Fig. 74 represent the cross-sections of two walls of equal 
height, and let the surface of the water be on a level with the top 
of each wall. 




260 


MASONRY DAMS AND RETAINING-WALLS. 


261 


II 


B 


E F 



Let d = depth of the water = height of the wall; 

RS = line parallel to AB and Id 
below it; 

a: = angle 1FH] 

w = weight of a cubic foot of 
water = 62.5 pounds; 

P' = resultant pressure on a 
linear foot of the back of 
the wall A BCD] 

P" = resultant pressure on a linear foot of the back of the 
wall EFGH. - 

From the principles above given, P', the resultant pressure 

on the back of a linear foot of the rectangular wall A BCD, will 
BD wd 2 

be wXBDX -=-. This resultant must act normal to the 

2 2 

surface BD, and at a distance § BD from B, or \BD from D. 

P", the resultant pressure on a linear foot of the back of 

d wd 2 

the trapezoidal wall EFGH, must be wXFHx— = -. 

2 2 cos a 

This resultant must act normal to the surface FH and at a distance 

%FH from F or \d from the surface of the water. 

The horizontal component of P" is 


wd 2 wd 2 

P'" = -Xcos a= = P\ 

2 cos a 2 


■ (481) 


Hence, whatever be the slope of the back of the wall, the hori¬ 
zontal component of the water pressure will be the same as the 
pressure on a wall of equal height with a vertical back, and will 
depend only upon the depth of the water. 

The vertical component of P" is 


piv __ 


wd 2 sin a wd 2 tan a 


2 cos a 


(482) 


. , . , d 2 tan a . . . , . . T TT , wd 2 tan a 

in which-is the area of the triangle BJH, and- 

2 2 

is the weight of a linear foot of the prism of water whose end is 

FJH. Its action line intersects GH at a distance \IH from 


















262 


CIVIL ENGINEERING. 


H. As the action line of the vertical component acts within the 
base of the wall, it increases the total pressure upon that base. 

The Profile.—The profile of the wall should fulfil the fol¬ 
lowing requirements: 

I. When the reservoir is full, the resistance of the profile 
should be sufficient to prevent the wall from shearing off in a 
horizontal plane, and the weight of the wall should be sufficient 
to prevent the wall from sliding on its base or on any horizontal 
joint. 

II. The resultant of the weight and the water pressure above 
any horizontal section, should pierce that section within the 
middle third, whether the reservoir be full or empty. This keeps 
the entire area of every horizontal joint in compression. 

III. The maximum pressure at the toe and heel of any joint 
should not exceed the allowable crushing value of the material 
of which the dam is constructed, nor should the maximum pres¬ 
sure at the toe and heel of the dam exceed the allowable bearing 
value of the material upon which the dam is built. 

In the discussion which follows it is assumed that the sur¬ 
face of the water is on a level with the top of the dam. 

Resistance to Shearing.—Resistance to shearing in masonry 
dams is secured by avoiding continuous horizontal joints and by 
bonding the wall with its foundation. Such a wall cannot rupture 
by simple sliding. 

Resistance to Sliding.—If the dam is constructed of hori¬ 
zontal layers of masonry, depending only upon the friction be¬ 
tween the layers to prevent sliding, stability requires that the 
friction at each joint shall be greater than the horizontal com¬ 
ponent of the water pressure upon the wall above the joint when 
the reservoir is full. 

If the dam has a vertical back and 

— weight of masonry per cubic foot = 144 pounds for ordinary 
masonry, 

w =weight of water per cubic foot = 62.5 pounds, 
h =height of wall and depth of water above the joint, 

5 = mean thickness of wall, 

/ = coefficient of friction of masonry on masonry = •§, 
for stability we must have 


MASONRY DAMS AND RETAINING-IVALLS. 


263 


wh 2 


wh 


w'bJi X/>-, .‘. b> —-.> 0.32 h. 

' 2 2W 7 


(483) 


Since b varies with h, a masonry dam of triangular profile, 
as shown in I and III, Fig. 75, is a dam equally strong at every 
horizontal joint to resist rupture by sliding. If in I, Fig. 75, 
BC> 0.64.AB, or b>o.^2h, the wall is stable at every horizontal 
joint. This is therefore the most economical profile for a dam 
with a vertical back to resist sliding, since the unit frictional re¬ 
sistance in every horizontal plane is the same. 

If the dam is constructed on an earth foundation, the base 
must be enlarged, because the coefficient of friction between 
masonry and earth is less than §. 

Masonry dams rarely fail because of weakness in horizontal 
planes, but-sometimes fail due to the insufficient frictional re¬ 
sistance at the foundation. 

The Middle Third. — In order that the action line of the 
weight shall intersect each horizontal section within its middle 
third when the reservoir is empty , the limiting forms of the pro¬ 
file must be a right-angled triangle with a vertical back, and a 
similar triangle with a vertical face, as in I and III, Fig. 75. 

1 n hi 



In I the action line of the weight alone passes through F, and in 
III through E. In the rectangular wall II the action line of 
the weight passes through F, the middle point of BC. 

Let it be required to compare the amount of masonry in 
the three walls of equal height, Fig. 75, under the requirement 
that the resultant of the maximum water pressure and the weight 
shall not intersect any horizontal section outside the middle 
third. This will be true in the sections shown when the result¬ 
ant pierces the base within the middle third. 


















264 


CIVIL ENGINEERING. 


Let h = height of dam = depth of water =AB; 
b = width of base = CB ; 

w'= weight of masonry per cubic foot = 144 pounds, as¬ 
suming specific gravity as 2.3; 
w = weight of water per cubic foot = 62.5 pounds. 

The effect of the water pressure in cases I and II is to cause 
the resultant to pierce the plane CB to the left of the point where 
it is intersected by the action line of the weight alone. In I and 
II the resultant may move from F, when the reservoir is empty, 
to E, when the reservoir is full; equating the moments of the 
weight and the water pressure about E , we shall have, 


w'hb b 11 w 2 

ini, - X- = - , wh 2 X—h i .*. b = h\l— , = o.66h =—h: (484) 

2 32 3 ^ w ' 3 


w 


b 11 \w 2 

in II, w'hbXT = —wh 2 X — h, .*. b = h\ —- = o.66h = -h. (48s) 
62 ^ >7*7 3 J 


In III, when the reservoir is full, the resultant must pass through 
E , where the action line of the weight intersects BC when the 
reservoir is empty; hence the moments of the vertical and hori¬ 
zontal components of the water pressure about this point must 
be equal, or 


whb b 1 , 1, 

-X — = —wh 2 X —h, 

2 3 2 3 


.*. b = h. 


(486) 


Under the above assumption, that the specific gravity of masonry 
is 2.3, the area of the profile of wall I will be \h 2 \ of wall II, \h 2 \ 
of III, \h 2 \ or in the proportion 2:4:3 approximately. 

The wall I is therefore the most economical form of the profile 
under the second requirement, page 262, since in any trapezoidal 
profile between I and II the areas would lie between those of I 
and II, and in any trapezoidal form between II and III the areas 
would lie between those of II and III. Its factor of safety against 
overturning about the toe is 2, or the moment of the weight about 
that edge is twice the moment of the water pressure. 

Distribution of Pressure on the Horizontal Joints. —Under 
the subject of eccentric loading it was shown that: 




MASONRY DAMS AND RETAINING-IVALLS. 


265 


(1) If the center of pressure due to the external forces, at any 
horizontal joint, is at the center of gravity, the pressure is uni¬ 
formly distributed; 

(2) If the center of pressure is at the extremity of the middle 
third, the pressure at the nearer edge will be twice the mean 
pressure, and at the farther edge zero; 

(3) If the center of pressure is without the middle third, the 
entire joint will not be under pressure, and the pressure at the 
nearer edge will exceed twice the mean pressure. 

Masonry dams should be so designed that the entire surface 
of each joint is under compression; hence the center of pressure 
should always be within the middle third. 

In applying these principles to the three profiles shown, let 
the common height of the dams be assumed as 60 feet, and the 
width of the base of I and II as §/z, or 40 feet. The weight of 
the dams per linear inch will be: I, 14,400 pounds; II, 28,800 
pounds; III, 21,600 pounds. The mean pressure on the base 
per square inch when the reservoirs are empty will be: 30 
pounds in I and III and 60 pounds in II. In I and III, since 
the center of pressure is at the extremity of the middle third, 
the maximum unit pressure will be 60 pounds and the minimum 
zero. In II the center of pressure is at the center of gravity and 
the pressure is uniform. Hence the maximum pressure due to 
weight of the dam alone is the same in the three profiles. 

When the reservoirs are full, the vertical component of the 
resultant pressure on dam I is the weight of the dam; it acts 
through E. The mean vertical pressure is therefore, as before, 
30 pounds; the maximum pressure at C is 60 pounds, and the 
minimum at B is zero. In dam II the vertical component of 
the resultant pressure is the weight of the dam; it acts through 
E, the extremity of the middle third. The mean pressure is 
therefore, as before, 60 pounds; the maximum at B is 120 pounds, 
and the minimum at C is zero. In dam III the vertical 
component is the weight of the wall and the vertical component 
of the water pressure, both of which act through E. The mean 

pressure is 30 +—30 = 43 pounds; the maximum pressure at 

B is 86 pounds, and the minimum pressure at C is zero. Hence 
profile I produces the least maximum pressure at the toe when 



266 


CIVIL ENGINEERING. 


the reservoir is full, and is the most economical profile under the 
third requirement, page 262. 

Modification of the Theoretical Profile.—Profile I, Fig. 75, 
has been shown to be the most economical profile under each 
of the three requirements of masonry dams; it should therefore 
be the basis of the designs for such structures. 

The upper part of dam I, Fig. 75, is designed to with¬ 
stand static water pressure alone. Dams of large reservoirs are 
subject to wave action and ice pressure, and overflow-dams to 
the pressure of the current and the shock of floating bodies. 
Upon the top of all except overflow-dams a footway at least three 
feet wide is desirable for communication, and upon all long 
dams a driveway at least 10 feet wide. The thickness of dams 
at the crest is therefore made from 3 to 20 feet, and the upper 
part of the profile is approximately rectangular. 

If a dam constructed as in I, Fig. 75, were 120 instead of 60 
feet high, since its weight increases with the square of the height, 
and its base as the first power of the height, the mean unit pres¬ 
sure at the base would be 60 pounds, and the maximum unit 
pressure would be 120 pounds; if 180 feet high, the mean unit 
pressure would be 90 pounds, and the maximum 180 pounds. 
It is not considered desirable in dams of moderate height to 
have the maximum unit pressure exceed 100 pounds per square 
inch, or about 7 tons per square foot; nor in very high dams 
is it considered safe to have a maximum unit pressure exceeding 
200 pounds per square inch or 14 tons per square foot. To 
limit the maximum pressures to the intensities above given it 
is necessary to increase the width at the base of all high dams 
beyond that given in I, Fig. 75. From this increase at the top 
and base of high dams, there results the form shown in Fig. 76, 
which is the profile of the new Croton Dam. In determining 
the resultant pressure upon such a dam the vertical component 
of the water pressure is usually omitted; as this component 
would tend to move the center of pressure towards the center of 
gravity when the reservoir is full, it is an error on the side of 
safety. The face of the dam is subjected to a back pressure 
due to the earth and water below the surface of the lower pool. 

Curve of Pressure.—The curve of pressure is the curve which 
connects the centers of pressure of the horizontal joints. To 


MASONRY DAMS AND RETAIN IN G-IVALLS. 


267 


find the curve of pressure of a dam, divide the profile by a num¬ 
ber of horizontal planes. When the reservoir is empty, the 
center of pressure at each horizontal joint is determined by drop¬ 
ping a perpendicular through the center of gravity of the section 



of the wall above it. When the reservoir is full, the center of 
pressure at each horizontal joint is the point where the resultant 
of the weight and the water pressure above the joint pierces that 
joint. 

To construct the curve of pressure of dam I, Fig. 75, let 


A = origin of coordinates; 


axis of X = AB ; 

axis of Y = line through A parallel to BC ; 
AH=x; 


GH = x tan <f>, in which (f> is the angle BAC ; 

IH = y = distance of center of pressure from AB when the 
reservoir is full; 

n/ —weight of unit volume of masonry; 
w= “ “ “ “ “ water. 


Equating moments of weight of wall and pressure of water 
about I, we have 




















268 


CIVIL ENGINEERING. 


y/x 2 tan cf> l x tan </>\ wx 3 

or 

y _ ^+?^l*=ik + kh. . . 

y 3 uj tan 9 3 

Similarly for dam II, Fig. 75, we have 

/ b \ wx? 
(w'xb)[y--j=—, 

in. which b is the breadth of the dam, or 

y=r'^ t+-=ik+kh . 

J 6 wb 2 


i n nr 





In dam III, Fig. 75, the sum of the moments of the weight of 
the wall above GH, of the water resting on the wall above GH , 
and the horizontal water pressure on AH must be equal to zero. 
Hence 


w'x 2 tan (j) ( x tan ^ wx 2 tan cf> (2 X tan <j> 

y ~ 3 


. wx 3 

-y) + ~6= °> 


or 


y= 


(w f 4- 2 w)x tan 2 (j) — wx 
3(w'+w) tan (f> 


(c) 


In the formulas (a), ( b ), and (c) the value of y for the empty 
reservoir may be obtained by making w = o. 

The curves of pressure resulting from the formulas (a), (b), 
and (c) are shown in dotted lines, Fig. 75. 

































MASONRY DAMS AND RETA IN IN G-IVALLS. 


269 


If a dam rests on a fissured or a permeable foundation, or 
the connection between the dam and foundation is not water¬ 
tight, there will be an upward pressure on the base of the dam 
which will greatly reduce the resultant vertical component of 
the pressure of the dam on its base. The upward pressure is 
avoided by constructing the dam upon solid rock, and by mak¬ 
ing the joints between the dam and its foundation water-tight. 

PROBLEMS. 

53. What should be the thickness of a rectangular masonry 
dam whose height is 30 feet if the level of the reservoir is 2 feet 
below that of the dam? Specific gravity of masonry is 2.3. Maxi¬ 
mum pressure must not exceed twice mean pressure. 

Ans. 17.8 feet. 

54. What will be the maximum and mean pressure upon the 
base of the dam in problem 53? 

Ans. Maximum 8640 pounds per square foot. 

Mean 4320 

55. A masonry dam with a vertical back is 48 feet high, 14 
feet wide at the top, and 28 at the base. Determine the position of 
the center of pressure of base when the dam is full. 

Ans. 9.16 feet from toe. 

Retaining-walls. 

A masonry retaining-wall is one designed to resist the lateral 
pressure of the earth upon one of its sides. The terms back, 
face, base, batter, etc., are applied as in masonry dams. If the 
earth is higher than the wall, the volume above the top of the 
wall is called the surcharge. In Fig. 77 the prism of earth MBE 
is the surcharge. 

Pressure of Earth.—The pressure of earth against the back 
of the wall differs from that of water in that the particles of earth 
cohere to each other with considerable tenacity, especially when 
in a moist condition, and free movement of the mass is resisted by 
friction. The cohesion is a very uncertain factor in determin¬ 
ing the pressure of earth, as it varies from the tenacity of com¬ 
pact clay, which when dry will stand at any angle, to that of 
dry sand, which is practically without cohesion. It varies in the 
same material with the degree of moisture and compactness, and 
is not uniform through any large mass of earth. 


270 


CIVIL ENGINEERING. 


To avoid these irregularities, formulas for earth pressure are 
based on the following hypothesis: 

I. The cohesion of the material has been largely destroyed 
and the earth consists of sandlike grains which are free to move, 
subject to the force of friction. It is evident that such a hypothe¬ 
sis is on the side of safety. 

If earth, in this condition, is poured out of a vessel on a smooth 
surface, it will form a cone the inclination of whose surface with 
the horizontal will vary slightly with the percentage of moisture. 
The inclination of the surface to the horizontal is called the 
natural slope of the material; the angle of inclination is called 
the angle 0) repose. Since a particle is held in equilibrium on 
the surface by the component of its weight acting parallel to 
the surface, and by an equal force of friction developed by the 
normal component, the angle of repose is also the angle oj fric¬ 
tion , and the tangent of the angle of repose is the coefficient oj 
friction of the material. The angle of repose of wet earth is 
about 30°, of very dry earth 38°, of moist earth 45 0 . In for¬ 
mulas it is usually assumed to be about 34 0 ; its natural tangent 
being §. 



Fig. 77. 


In Fig. 77 A BCD is a section of a straight retaining-wall; 
BF is the upper surface of the earth in its rear, which is assumed 
to be horizontal; DF is the natural slope of the earth. To 
deduce the formula for this pressure it is necessary to make 
further hypotheses. The two hypotheses ordinarily accepted 
are: 












MASONRY DAMS AND RETA IN IN G-IVALLS. 


271 


II. If the wall is suddenly removed, a portion of the volume 
BDF will at once fall; the remainder will assume its natural 
slope gradually. The pressure upon the back of the wall is 
assumed to be due to the volume which will at once fall, and 
this volume is assumed to be separated from the remainder by 
a plane surface DE, called the plane oj rupture, along which it 
slides in the form of a compact prism. The angle BDE is called 
the angle oj rupture. The less cohesion there is in the material, 
the more closely will the results of the experiment agree with 
this hypothesis. 

III. The pressure due to the volume BDE is distributed over 
the back of the wall BD, in the same manner as water pressure; 
the unit pressure at any point varies with its distance from B, 
and the center of pressure on a rectangular area whose height 
is BD is | BD from B. 

Coulomb’s Formula.—This formula is based upon the three 
hypotheses above given, and upon the additional one: 

IV. The action line of the resultant pressure upon the back 
of the wall is normal to the surface, as in water pressure, or the 
friction on the surface BD may be omitted. 

In Fig. 77 let 

h = height of the wall = BD\ 
angle <j> = angle of repose of the earth = FDG; 
angle a = angle of rupture of the earth = BDE; 
angle a' = EDF; 

w = weight of a cubic foot of earth; 

P = resultant pressure on BD ; 

W = weight of prism BDE per foot of length; 

p l= normal reaction of surface BD = P; 

R 2 = normal reaction of surface DE; 

R 3 = frictional resistance of surface DE; 

R4 = resultant of R2 and R 3 . 

Under the hypotheses above given there must be equilibrium 
between the weight of the prism BDE, the normal reaction of 
the surface BD, the normal reaction of the surface DE, and 
the friction on the surface DE caused by the weight of the prism. 
The weight of the prism BDE per foot of length is \wh 2 tan a; 
its action line is W, through its center of gravity and 0 . Of 


272 


CIVIL ENGINEERING. 


the other forces we know the action lines of R u R 2 , and R3 , 
but we do not know their intensities. We do know, however, 
that the friction on DE or R 3 must be equal to the normal 
pressure or normal resistance R 2 multiplied by the coefficient 
of friction, or R 2 tan </>. Hence if we draw R± making an 
angle </> with R 2 , it must be the action line of the resultant of the 
friction R3 and the normal reaction R 2 , whatever be their 
intensities. There must therefore be equilibrium between W } 
R 1, and R±, and our system is reduced to a system of three con¬ 
current forces of which we know the intensity of one, and the 
action lines of all three. We may therefore determine their 
intensities by constructing the force polygon HIJ, in which 
HI = W, IJ = Ri = P, and JH = R±. We may also determine 
the intensity of R 2 and R3 by drawing the lines JK and 
KH. Since the angle which W makes with R 2 = EDG, and 
the angle R 2 OR± = FDG , the angle IHJ = EDF=a f ; hence 
P = ^wh 2 tan a tan a'. 

This expression is a maximum when a=ct'; this may be 
shown by calculus. Hence the maximum value of P = \wh 2 
tan 2 a. 

I 7Z 

Since a=a'=— (90 — <£), and 90 =—, this may be also written 

. 2 2 

P=-wh? tan*g-f),. (488) 

which is the form in which it was originally deduced. 

By the same hypotheses we may deduce a general formula 
for the pressure of a surcharged embankment upon a wall with 
an inclined back; the formula is, however, very complicated. 

Poncelet’s Formula. —Poncelet takes into consideration the 
friction on the surface BD, and hence the resultant reaction of 
the surface BD is R 5 (Fig. 77), which makes with the normal an 
angle <f> f , equal to the angle of friction of earth on masonry. For 
all practical purposes <£' = <f>. Poncelet’s formula for the pressure 
upon a retaining-wall with a vertical back with the surface of 
the earth level with the top of the wall, in terms of is 

wh 2 cos (j) 

E5 P 1— 7 

2(1 + V2 sin ( j >) 2 


■ ■ (489) 



MASONRY DAMS AND RETA IN IN G-IVALLS. 


273 


Rankine’s Formula.—This formula is based upon the theory 
of internal stresses in a granular mass. The conclusions he 
arrives at are that the pressure at any point of BD varies with 
the distance from B, and that the action line of the resultant of 
the pressure is parallel to BM whenever BM makes an angle 
with the horizontal varying between zero and <f>. If BM is 
horizontal, Rankine’s formula is 

1 — sin <f> 

1 + sin cf> 

If BM is parallel to DF, this formula becomes 

n wh 2 cos cj) 

£ • 

2 

More general forms of this formula are given by Rankine. 
Other theoretical formulas based on the hypotheses above given 
have also been deduced. 

The earth pressure on a retaining-wall may also be deter¬ 
mined by the graphic method. While no simpler than the 
analytic method, it may be utilized to check results obtained 
by calculation. The methods are given in works on graphic 
statics. 

Having determined the earth pressure upon a retaining-wall, 
the form of cross-section is designed upon the same principles as 
that of a masonry dam. 

I. The horizontal width of the profile at every point should 
be sufficient to prevent the wall from shearing off in any hori¬ 
zontal plane or sliding on its foundation. Walls laid without 
mortar are liable to rupture by shearing or bulging. 

II. The resultant of the weight and the earth pressure should 
pierce each horizontal section far enough within the outer edge 
to make the wall safe against overturning. This is accomplished 
by making the moment of the wall about this edge twice as great 
as the moment of the earth pressure, or, preferably, by limiting 
the curve of pressure to the middle third of the wall. 

III. The maximum unit pressure at the toe and heel should 
not exceed the bearing value of the foundation, nor should the 
maximum unit pressure on any joint exceed the allowable com¬ 
pressive stress of masonry. 




1 





274 


CIVIL ENGINEERING. 


As a retaining-wall is usually constructed to support an em¬ 
bankment of considerable height, in designing the profile it is 
only necessary to consider the pressure of the earth when the 
embankment is in its final condition. 

The weight being the same, the most stable wall to resist 
overturning is therefore the one in which the center of gravity 
is at the greatest distance from the toe. 

Poncelet deduced the following formula for converting a rect¬ 
angular wall into a trapezoidal wall with a battered face, vertical 
back, and an equal moment about the toe. Its accuracy in¬ 
creases with the inclination of the face. 


b ' = b + i \ b ", .(492) 


in which b'= base of trapezoidal wall; 

b =base of rectangular wall; 


b " = assumed base of battered face = 


h 


tan 


h = height of wall; 

/? = angle of inclination of face to the horizontal. 

A profile in the form of a parallelogram 
leaning towards the embankment offers much 
greater resistance to overturning than a rect¬ 
angular wall having the same base and height. 
Such a wall would, however, be apt to slide 
upon its base. 

The most common form of retaining-wall. 
Fig. 78, is one which has a vertical or slightly 
battered face and a back made in rough 
steps. The top width of a retaining-wall is usually about 2 
feet. 

Empirical Formulas.—Although the theoretical formulas for 
earth pressure will give safe values for the thickness of retaining 
walls, their forms are very complex for surcharged walls and for 
wails with inclined backs. For these reasons and because they 
are based on hypotheses only approximately true, empirical 
formulas have largely replaced them. 

Let h = height in feet of wall of rectangular cross-section; 

6 = mean thickness in feet of wall of rectangular cross- 
section. 










MASONRY DAMS AND RETAINING-WALLS. 275 

Trautwine recommends the formulas for walls without sur¬ 
charge 

b = 0.35 h for first-class masonry.(493) 

6 = 0.40/^ “ rubble masonry laid in mortar . (494) 
5 = 0.50/^ “ well laid dry rubble.(495) 

French engineers use the formula 

b = 0.7,0k for first-class masonry.(496) 

A wall with a rectangular cross-section can be transformed 
into one of equal stability with a battered face and vertical back 
by means of formula (492). 

If the face is vertical and the back is stepped the formulas 
above given may be employed, the mean thickness of the vertical 
profile of the wall will be about 0.85 of the value of b, given 
above. 

If the face or back of a wall is not vertical, the top should be 
at least two feet thick. 

Surcharged Walls.—A surcharged wall must be made thicker 
than one without a surcharge. 

If H = height of surcharge in feet, or vertical height of M 
(Fig. 77) above AB, the French engineers’ formulas for sur¬ 
charged walls of rectangular cross-section are 

b=\h+\H if H <—, 

b=%h+^ z H if H > 2b . (497) 

For intermediate values of H in terms of h the fractional 
coefficients of H in the formula are obtained by interpolation 
between J and 

Foundations.—The empirical rules apply only to walls on 
firm foundations. The failure of masonry retaining-walls laid 
in mortar is usually due to defects in the foundations which cause 
the walls to slide on their foundation or to settle unequally. This 
is especially true of retaining-walls along waterways where the 
soil is of a soft, compressible character. 

Drainage.—If the earth in rear of a retaining-wall becomes 
saturated with water, it will greatly increase the pressure upon 



276 


CIVIL ENGINEERING. 


the back of the wall. To avoid such increased pressure a verti¬ 
cal layer of broken stone is placed against the back of the wall, 
and drainage-tubes, or weepers, are run through or under the 
wall, to allow the escape of the water. 

Counterforts and Buttresses.—A counterfort is a projection 
upon the back of the wall designed to strengthen the wall to 
resist the pressure of the earth; a buttress is a similar projection 
on the face of the wall. Counterforts of simple masonry are of 
doubtful efficiency in strengthening a wall either against shear¬ 
ing or overturning because of their liability to separate from the 
wall itself. A better wall is usually secured by placing all the 
masonry in the wall itself. They are sometimes employed in 
long walls to divide it into panels, and have been largely em¬ 
ployed in the masonry walls of fortifications to limit the field 
of destruction of an exploding shell or mine and to form the side 
walls of casemates. In walls of concrete constructed about a 
framework of steel, counterforts may be constructed which are 
an integral part of the wall and cannot be separated from it. 

Buttresses are more efficient than counterforts in strength¬ 
ening a wall against rupture by shearing or overturning. They 
are not often employed for this purpose except in Gothic archi¬ 
tecture, as it is usually desirable to have the face of the wall a 
plane surface. 


PROBLEMS. 

57. Determine by the use of Coulomb’s Formula the thickness 

of a rectangular retaining-wall 20 feet high to safely resist the 
pressure of an embankment of equal height. Specific gravity of 
earth 1.5. Specific gravity of masonry 2.3. Angle of repose of 
earth =34°. Make the moment of the weight of the wall twice that 
of the earth pressure. Ans. 7.0 feet. 

58. Solve problem given above by Poncelet’s Formula. 

Ans. 6.1 feet. 

59. Solve the problem given above by Rankine’s Formula. 

Compare the thickness with that obtained by using empirical 
formula b =0.4 h. Ans. 7.0 feet. 


CHAPTER XVI. 


MASONRY ARCHES. 


Definitions.—A masonry arch (Fig. 79) is a structure designed 
to support pressure and transmit its effects to lateral points of 



support. The arch is ordinarily composed of wedge-shaped 
blocks called voussoirs , which support each other by lateral pres¬ 
sure. 

Soffit. —The inner cylindrical surface of the arch. 

Back. —The outer surface of the arch. 

Intrados. —The line of greatest curvature or the curve of 
right section of the soffit ( ACB ). 

Extrados. —The line of greatest curvature or the curve of 
right section of the cylinder containing the outer extremities of 
the joints ( EFG ). 

Abutment. —The mass of masonry designed to resist the 
thrust of the arch (. AELKH ). It is often a retaining-wall. 

Pier. —A column of masonry designed to resist only the verti¬ 
cal component of the thrust of two adjacent arches (. BGBJI ). 

Skewback. —The joint, usually inclined, between the extreme 
voussoir and the abutment or pier (EA). 

277 

















2 jS 


CIVIL ENGINEERING. 


Cushion-stone. —The stone whose upper surface is the skew- 
back {A EL). 

S pringing-line. —The intersection of the skewback and the 
soffit (A). 

Span. —The perpendicular distance between the springing- 
lines ( AB). 

Crown. —The highest rectilinear element of the soffit (C). 

Rise. —The vertical distance between the crown and the plane 
•of the springing-lines {CD). 

Axis. —The intersection of the plane of the springing-lines 
by a vertical plane through the crown {D). 

Keystone. —The highest stone of an arch (VI). 

Springer. —The lowest stone of the arch (II). 

Head. —The end surface of the arch. 

Haunches. —The part of the arch about midway, vertically, 
between crown and springing-lines (III, IV). 

String-course. —A row of voussoirs parallel to the axis. 

Coursing-joint. —The joints between string-courses (ON). 

Ring-course. —A ring of voussoirs parallel to the head of an 
arch. 

Heading-joint. —The joint between the ring-courses. 

Spandrel. —The volume above the arch limited by the back 
of the arch, vertical planes through the outside edges of the 
skewbacks, and a horizontal plane parallel to the plane of the 
springing-lines (EPQGF). A spandrel wall is one constructed 
in this space, parallel to the head of an arch; spandrel filling is 
the material deposited in the spandrel. 

Form of Intrados.— Full-tenter Arch. —One whose intrados is 
a semicircle. 

Segmental Arch. —One whose intrados is a circular arc less 
than i8o°. 

Elliptical Arch. —One whose intrados is a semi-ellipse. 

Oval Arch. —One whose intrados is an oval or curve similar 
to an ellipse made up of three or more circular arcs tangent to 
each other. 

Tudor Arch. —A pointed arch whose intrados is made up of 
two intersecting curves. 

Direction of Axis.— Right Arch. —One in which the axis is 
perpendicular to the head. 


MASONRY ARCHES. 


2 79 


Oblique or Skew Arch. —One in which the vertical plane 
through the axis makes, an oblique angle with the plane of the 
head. 

Rampant Arch. —One in which the axis is inclined to the 
horizontal plane. 

Groined and Cloistered Arches. —The soffits of groined and 
cloistered arches are formed by the intersection of two cylin¬ 
drical soffits having the same rise and intersecting axes. In a 
groined arch that part of each soffit which lies within the other 
is removed, thus preserving the crown of each arch throughout. 
The groins are the curves of intersection of the two soffits. 
The cloistered arch is a dome-shaped arch which rises from 
the springing-lines to a point. It is formed by retaining only 
that part of each soffit which lies within the other. 

Annular Arch. —An annular arch is one generated by revolv¬ 
ing a plane of right section about a line of this plane perpendicular 
to the span, but not intersecting the arch; it is a circular arched 
passageway. 

Dome. —The soffit of a dome is generated by revolving a plane 
of right section about a vertical line intersecting its crown. 

Theory of the Arch. 

Modes of Rupture.—An arch, like a masonry dam, may 
rupture by rotation, by sliding, or by the crushing of the material. 
The mode of rupture depends upon the form of the arch and 
the depth of the voussoirs. 

The ordinary mode of rupture of a full center arch, in 
which the ratio of the rise to the span is one-half, is shown 
in I, Fig. 80; the thrust of the arch at the haunches causes the 



lower segments to rotate outwards about the outer edge of the 
skewback or of some joint of the abutment, and when the points 
F and F' are sufficiently separated, the weight above the crown 




280 


CIVIL ENGINEERING. 


causes the upper segments to rotate inwards. An oval or ellip¬ 
tical arch may also rupture as shown in II, Fig. 81. The thrust 
at the haunches causes the lower segments to slide on the cushion- 
stones and thus separate the points F and F when these are 
sufficiently separated the upper segments rotate inwards as 
before. In the arches above described, the points F and F r lie 
in a horizontal plane which passes midway between the crown 
and the horizontal diameter of the circle or the horizontal axis 
of the ellipse or oval. If this plane passes below the springing- 
lines of a segmental arch, the points F and F r are at the spring¬ 
ing lines. A segmental arch usually ruptures as shown in I, Fig. 
81. The thrust at the springing-lines causes the arch to slide 



Fig. 8i. 


on its abutments and separates the points F and F' sufficiently 
to allow the voussoirs near the crown to slip through. 

The ordinary mode of rupture of an arch in which the ratio 
of the rise to the span exceeds one-half is shown in II, Fig. 80. 
The pressure on the haunches causes the arch to open at some 
joint, as EF, near the crown. The lower segment then rotates 
inwards about some joints, as CD, near the springing-lines, 
thus bringing the points F and F' nearer together; as the points 
E and E' move, they cause the crown of the arch to rise and the 
upper segments to rotate about B. 

The joints shown in Figs. 80 and 81, at which the arch opens, 
are called the joints oj rupture. 

If the arch does not rupture by rotation or by sliding, it can 
rupture only by the crushing of the material of which it is com¬ 
posed. If the arch ruptures in this manner, it will crush the 
material at the joints of rupture, since, as will be shown, the’ 
resultant pressure is nearer the intrados or extrados at these 
points than at any other points of the arch. 


/ 










MASONRY ARCHES. 


281 


The Thrust at the Crown.—Let Fig. 82 represent an arch 
made up of two monolithic segments with a vertical joint at the 
crown. Assume the arch to be made of material of infinite 
strength and without elasticity. Each semi-arch will, under the 



action of its own weight, attempt to rotate about its springing- 
line and thus produce a thrust upon the other half of the arch. 
As in a three-hinged truss, each semi-arch, when at rest, is acted 
upon by a system of three forces in equilibrium, consisting of 
W 

its weight —, which acts through its center of gravity, the thrust 
2 

at the crown H , or pressure of the other half of the arch, and 
the reaction at the skewback R r A system of three forces in 

W 

equilibrium must either be a parallel or a concurrent system. — 

and H cannot be parallel since one is a vertical force and the 
other is the reaction of two vertical surfaces and cannot be verti¬ 
cal. The system is therefore a concurrent one. Such a system 
admits of accurate solution if one of the forces is fully given 
and the action lines of the other two are known. 

W 

Of the three forces, the intensity and action line of — can 

2 

be determined when the dimensions of the arch and the specific 
gravity of the material are known. If the two semi-arches are 
symmetrical and are symmetrically loaded, the action line of H 
will be horizontal. The point of application of H may, how¬ 
ever, be anywhere between A and B. 

Of the force R / we know only that its point of application 
must lie between C and D. An accurate solution of the system 
is therefore impossible. We may, however, determine the limits 
within which the intensities of H and R / must lie. 


1 











282 


C/K/I ENGINEERING. 


We may assume that H acts at A or B , and that R , acts at C 
or 7 ). 

These hypotheses will give the limiting values of H and R,. 

Let H' = intensity of H when the point of application is at 
A and the center of moments is at 77 ; 

H" = the intensity of H when the point of application 
is at A and the center of moments is at C; 

H" r = intensity of H when point of application is at B and 
center of moments is at D ; 

jE 7 iv =intensity of H when point of application is at B 
and center of moments is at C; 

y" = lever-arm of H with respect to D or C when point 
of application is at A ; 

y = lever-arm of H with respect to D or C when point 
of application is at B ; 

W 

xf = lever-arm of — with respect to D\ 

*2 

w 

x" = lever-arm of — with respect to C . 

Then we shall have: 

Point of application A : 


Minimum value H' = 


Wx r 


2 y n 


Wx" 

Maximum value H" = — -77-, 

27' 


Point of application B : 


Minimum value H" f = 


TIV 


27 


/» 


Maximum value H lv = 


Wx" 


2J 


To interpret these values make AB in II, Fig. 82, equal to AB 
in I, Fig. 82. 

At the extremity of AB in II, Fig. 82, lay off Aa = H ', Ab = H ", 
Ba = H"', and Bb = H ly . Then is aa the curve whose ordinates 
give values of H y which, if applied at the proper points of the 





MASONRY ARCHES. 


283 


w 

joint AB, will make the resultant of — and H act through D, 

and bb is the curve whose ordinates will give the values of H, 

W 

which, if applied at the joint AB, will make the resultant of — 

2 

and H act through C. The curves aa and bb are hyperbolas 
since Hy is constant. 

The area aabb may be taken as a measure of the stability of 
the arch, since the ordinate of any point of the surface will give 
a value for H which, if applied at the proper point of AB, will 
prevent the semi-arch from rotating about D or C. Any value of 
H less than the ordinate of aa, or greater than the ordinate of bb, 
will cause the semi-arch to rotate about D or C. Thus the 
intensities Ga, Gb, or any intermediate values, will, if applied at G 
of the crown joint, prevent the arch from rotating about D or C. 
In the actual design of an arch some latitude must be al- 

W 

lowed for errors in the value of H, due to the fact that — can- 

2 

not be accurately determined. Furthermore, if the material is 
of finite strength, the thrust cannot act at A or B without crush¬ 
ing the material, since the unit pressure, as was shown under 
eccentric loading, would then be infinite. For these reasons, 
the hypothesis is made that H and R y shall act within the middle 
third of their respective joints; the limiting points of application 
of H in I, Fig. 82, are therefore G and H; and of R / E and F , 
in which GH = \AB, and EF = \CD. 

The limiting values of H consistent with satisfactory sta¬ 
bility are shown in Fig. 82, in which 


JK=NO = 


IV 


KM and KL = the maximum and minimum values of H 
when the point of application is at G. 

OQ and OP = the maximum and minimum values of H when 
the point of application is at H. 

If in II, Fig. 82, we lay off AG = GH=HB, Gc = KL, Hc = 
OP, Gd = KM, Hd = OQ, the area ccdd may be taken as the 
measure of the satisfactory stability of the semi-arch. The 



284 


CIVIL ENGINEERING. 


ordinate of any point of this area will give a value of H which, 
acting at the corresponding point within the middle third of 

W 

AB, will cause the resultant of — and H to intersect CD be- 

2 

tween E and F. 

If the semi-arch is designed by employing such a value of 

H, the arch cannot yield by rotation under the crown thrust; it 

may, however, slide on the skewback if the action line of the 

W . . . 

resultant of — and H is sufficiently oblique to the joint at the 
2 

springing-line. In order that sliding shall not take place, the 
force of friction at this joint must be greater than the compo- 

W 

nent of the resultant of — and H parallel to the joint, or, in other 

2 

words, the resultant must make a less angle with the normal 
to the joint than the angle of friction. 

Assuming the angle of friction of masonry on masonry to be 
about 37 degrees, the angle between the actual resultant and 
the normal must be less than this angle. As the action line of 
the actual resultant cannot be accurately fixed, the angle between 
its most oblique action line possible, NS, Fig. 82, and the normal 
is, for safety, made less than 37 degrees. 

If the two conditions as above given are fulfilled, the arch 
cannot rupture unless the masonry itself crushes. To prevent 
this, the unit pressure at A, B, C, and D should not exceed the 
allowable unit compressive stress of the masonry of which the 
arch is composed. 

General Condition of Stability.—If we now assume the semi¬ 
arch to be divided into voussoirs, it is evident that the same con¬ 
ditions which we have imposed upon the semi-arch must be im¬ 
posed upon each segment included between the crown and any 
joint; hence we have the following conditions of satisfactory 
stability for an arch. 

1. The resultant 0} the thrust at the crown and the pressure 
upon the arch between the crown and any joint should make a 
less angle with the normal to the joint than the angle oj friction. 
This will prevent sliding at the joints even if no reliance is 
placed upon the strength of the mortar. 

2. The resultant of the thrust at the crown, and the pressure 


MASONRY ARCHES. 285 

upon the arch between the crown and any joint , should pierce the 
joint within the middle third. 

3. The unit pressure at the extrados or intrados oj any joint 
should not exceed the allowable unit stress in compression oj the 
material oj which the arch is made. 

Mery’s Curve of Pressure.—If we know the intensity and 
point of application of the thrust at the crown, it is an easy mat¬ 
ter to construct graphically the resultant of the thrust at the 



crown, and the weight of each segment, and find the correspond¬ 
ing center of pressure. Thus in Fig. 83 let 

DC = IF' — action line and intensity of weight of vous- 
soir I; 

GF = IF' + IF" = action line and intensity of weight of vous- 
soirs I and II; 

IF 

JI = W' + W " + IF'" = — = action line and intensity of weight 

2 

of voussoirs I, II, and III; 

AB = action line of thrust at crown; 

JK = GH = DE = intensity of the thrust at the crown. 

Combining IF' and H , the resultant is EC , which pierces the 
joint between I and II at L. 

Combining IF' + IF" and H, their resultant is HF, which 
pierces the joint between II and III at M. 

Combining IF' + IF" + IF'" with H , their resultant pierces the 
skewback at N. 






286 


CIVIL ENGINEERING . 


If we now construct a curve passing through BLMN tan¬ 
gent to the several resultants, it will be Mery’s curve oj pressure 
or the curve oj resistance. 

If the curve of pressure of an arch is given, its stability can 
be readily tested. If at any joint the angle between the normal 
to the joint and the tangent to the curve of pressure is less than 
the angle of friction, the first condition of satisfactory stability 
is fulfilled. If the curve of pressure lies everywhere within the 
middle third, the second condition of satisfactory stability is 
fulfilled. 

The curve of pressure may also be constructed by means of 
an equilibrium polygon. 

IF 

In Fig. 83 lay off Oa equal to JK or iT, and ab equal to — or 

IF' + IF" + IF'". Divide ab into three parts, equal to the weights 
IF', IF", IF"', and draw lines 1, 2, and 3. From the force poly¬ 
gon Oab construct the equilibrium polygon BO'O n O'"Q. Then 
will the points where the sides of the polygon intersect the joints 
be the centers of pressure, and a curve tangent to the equilib¬ 
rium polygon at these points will be the curve of pressure. 

The equilibrium polygon is called the line oj pressure since 
its lines are the action lines of the pressures at the joints. 

It is evident, from the discussion given above, that the curve 
of pressure of a symmetrically loaded arch may always be con¬ 
structed when the intensity and the point of application of the 
crown thrust are known. 

In the segmental arch shown in Fig. 83 the theory of mini¬ 
mum crown thrust has been adopted. According to this theory 
the maximum pressure at any joint must not exceed twice its 
mean pressure, and the intensity of the crown thrust is the mini¬ 
mum thrust which will fulfil this condition. 

In Fig. 83 the crown thrust, according to this theory, must 
be ED = HG = KJ , and its point of application must be B. If 
a thrust of less intensity than ED acts at B, the center of pressure 
at the joint BO will lie between N and O, or be outside of the 
middle third, and the pressure at O will be greater than twice the 
mean pressure on PO. If the point of application of ED is 
above B, the pressure at the top of the crown joint will be greater 
than twice the mean pressure on the crown joint. If the point of 


MASONRY ARCHES . 


287 


application of ED is below B , the center of pressure at the joint 
PO will lie between N and O, or outside the middle third, and 
the pressure at O will be greater than twice the mean pressure on 

PO. 

The minimum crown thrust of a symmetrically loaded arch 
may be readily determined if the joints of rupture are known. 
In a segmental arch, like that shown in Fig. 83, we know that 
the skewback PO is a joint of rupture. If, therefore, we con¬ 
nect /, the point of intersection of the action lines of the crown 
thrust and the weight of the semi-arch, with N, the extremity 
of the middle third of PO, it will be the resultant line of pressure 
of the weight of the semi-arch and the minimum crown thrust. 
If we lay off IJ equal to the weight, KJ must be the intensity 
of the minimum crown thrust. 

W 

Load on an Arch.—In determining the value of — for the 

semi-arch and the value of IF', IF", etc., for each voussoir, we 
must consider not only the weight of the voussoirs themselves, 
but also the pressure to which they are subjected from the span¬ 
drel filling. In determining these values it is usual to reduce the 
area of the spandrel filling to an equivalent area having the same 
specific gravity as the masonry of the arch. In the arch shown 
in Fig. 84 the ratio of the specific gravity of the spandrel filling 



to that of the masonry is assumed to be §. We may therefore 
reduce the spandrel volume to one of the same specific gravity 
as the masonry by laying off from the back of the arch on each 
ordinate of the spandrel, as EF, a distance EG = %EF. The line 
LM connecting these points will be the cross-section of the upper 






















288 


CIVIL ENGINEERING. 


surface of the spandrel filling having the same specific gravity as 
the masonry of the arch. 

To find the weight resting on each voussoir approximately, 
the spandrel filling is divided by vertical planes, as shown in 
Fig. 84. The voussoir is assumed to form a single mass with 
the load upon it, hence its weight is assumed to act througli the 
center of gravity of the mass. 

To find the action line of the weight of any of these masses, 
as PEGIJK , find the area of surface EGIJ assumed as a rect¬ 
angle whose width is HF and depth NO, and its moment with 
respect to any point, as C, the crown of the arch; find also the 
area of the voussoir EPKJ, assumed to be a trapezoid, and its 
moment with respect to the same point. The sum of the moments 
divided by the sum of the areas will give the approximate dis¬ 
tance of the center of gravity of the mass from C, and a 
vertical line at this distance from C will be the action line of 
its weight. 

In the above discussion we have neglected the friction along 
the planes HJ and FE , and have assumed that the block KJIGEP 
is a single mass. It would probably be more correct to as¬ 
sume the weight of the block JIGE as acting vertically through 
O, and at that point resolve it into normal and tangential com¬ 
ponents. 

Depth of Keystone.—If the curve of pressure lies within the 
middle third of the joint, the unit pressure upon the extremity 

2 H 

of the keystone joint cannot exceed —-j~, in which H equals the 

horizontal thrust in pounds of an arch one inch in length, and 
d equals the depth of the keystone in inches. H, however, cannot 
be known until the form of the right section of the arch is known. 

In designing an arch the depth of the keystone must be as¬ 
sumed. Various formulas have been proposed by engineers for 
this purpose, usually based upon a study of arches already con¬ 
structed and assumed to be satisfactory. As H is dependent upon 
the span and rise, these quantities must enter a general formula. 
In the following formulas for circular arches the dimensions are in 
feet; d = depth of the keystone, h = rise of arch, 5 = span of arch. 

Perronet: 


d= 1 tV feet -baV 5 full center arch. . . 



MASONRY ARCHES. 


289 


This formula makes d too great when s > 50 feet. 
Dupuit: 


d = 0.13s for full-centered arches,.(499) 


d = 0.073s for segmental arches when 


h 1 
s ~4 * 


(500) 


Dupuit’s formulas are recommended by the Manual of the 
“Ponts et Chaussees,” France. 

Dejardin: 

d — 1 foot + 0.05s when — = —, . . . (501) 


d— 1 


foot + 0.026s when 



(5°2) 


The coefficient of s is interpolated when — varies between 

s 

2 and yq-. 

Depth at the Springing-line. —In order that the mean pressure 
shall be the same throughout, the arch must increase in depth 
from the crown to the springing-lines. In arches of large span 
the ratio of the depth at the keystone to the depth at the spring¬ 
ing-lines varies from § to J. In arches of small span the depth 
is usually constant. 

Allowable Masonry Pressures. —The following are considered 
safe limits for the allowable unit pressures in arches: 

Concrete, 60 to 70 pounds per square inch. 

Brick in lime mortar, 85 pounds per square inch. 

Brick in hydraulic cement mortar, no to 114 pounds per 
square inch. 

Soft stone masonry in hydraulic cement mortar, 85 to 200 
pounds per square inch. 

Hard stone masonry in hydraulic cement mortar, 300 to 400 
pounds per square inch. 

Testing the Design of an Arch. —The curves of the intrados 
and extrados, the depth at the crown and springing-lines, the 
specific gravity of the masonry, the specific gravity and form 
of the spandrel filling must all be assumed. From the data thus 
given the load upon every voussoir is determined. 


290 


CIVIL ENGINEERING. 


If the coursing-joints are made normal to the intrados, there 
will be little danger of the voussoirs sliding upon each other, 
since the resultant of the external forces acting on either side of 
the joint will make with the normal a less angle than the angle 
of friction. Rupture, as shown in I and II, Fig. 81, is prevented 
by strengthening the abutment. 

In this manner the first condition of stability is fulfilled. 

The second and third conditions of stability are fulfilled by 
making the distance between the intrados and the extrados 
sufficient at every point to cause the center of pressure to lie 
within the middle third of every joint. 

If the arch is designed by the theory of minimum crown 
thrust, we may always determine the intensity of this thrust in 
a symmetrically loaded arch if we know the joints of rupture. 
If the arch is a full-centered one, as Fig. 80, we know its joints 
of rupture are EF and CD. The minimum crown thrust for 
the joint EF may be determined, as explained for the segmental 
arch, by assuming that at the joint EF the center of pressure is 
one-third of EF from F; if this thrust will cause the center of 
pressure at CD to lie within the middle third, the arch will have 
satisfactory stability. If the minimum crown thrust as above 
determined causes the center of pressure at CD to lie outside 
the middle third, the thickness of the arch at CD must be increased 
until the center of pressure lies within the middle third. It is 
usually done by moving C farther from D and making the 
extrados a straight line from the haunch to the springing-line. 

If the joints of rupture are not known, we may construct for 
each joint the limits Gc and Gd, Fig. 82, within which the mini¬ 
mum crown thrust must lie for that joint; values common to 
all joints may be assumed as satisfactory values of the crown 
thrust to be employed in designing the arch. 

If the arch shown in Fig. 82 is designed with tight joints and 
with a minimum crown thrust Gc, it is evident that unless there 
is some yielding in the material the actual minimum thrust may 
be less than Gc and act between G and A. If the crown thrust 
acts at any point between G and A, the maximum pressure will 
exceed twice the mean pressure. We can therefore insure com¬ 
pliance with the condition that the maximum shall not exceed 
twice the mean pressure only by making the joint open between 


MASONRY ARCHES. 


291 


G and A. In some segmental arches recently constructed, lead 
plates were inserted at the crown and springing-lines, covering only 
the middle third of the arch, and the remainder of the joint was 
filled with a non-resisting material. In such an arch the maxi¬ 
mum pressure at a joint cannot exceed twice the mean pressure. 

Elastic Arch.—The arch, especially if a monolithic structure 
of concrete, may also be designed on the theory that it is a curved 
beam whose fibers have tensile and compressive resistance.* 

Abutments. — The abutment should fulfil conditions similar 
to those of the arch itself. If the abutment resists by its weight 
alone, the resultant of the thrust of the arch at the skewback, 
and the weight of the abutment above any horizontal section, 
should pierce the section within its middle third. The abut¬ 
ment should also have sufficient strength in every horizontal plane 
to resist the horizontal component of the thrust. 

In computing the dimensions of the abutments the line of 
pressure should be assumed in its most unfavorable position for 
the stability of the abutment. This is usually through the crown 
itself and the outer edge of the skewback. 

Unsymmetrical Loading.—If the arch is not symmetrically 
loaded, the crown thrust will not be horizontal. Its direction 
may be determined either by assuming points of the curve of 
pressure at the skewbacks and constructing an equilibrium 
polygon passing through them, or by assuming points of the 
curve of pressure both at the skewbacks and crown, and 
constructing an equilibrium polygon passing through the three 
points. In either case it is necessary to experiment until a poly¬ 
gon is constructed which intersects every joint within its middle 
third, or it is clear that such a polygon cannot be constructed. 
In the latter case the dimensions of the arch must be changed. 
In determining the effect of a live load it is customary to assume 
a uniformly distributed live load extending from the crown to 
one of the abutments. 

The line of the equilibrium polygon intersecting the crown 
joint will be the action line of the crown thrust for unsymmet¬ 
rical as well as for symmetrical loading. 


* “ A Treatise on Arches,” by Malverd A. Howe. 




CHAPTER XVII. 


THE PRESSURE AND FLOW OF WATER. 

The design of the structures employed in systems of water- 
supply and sewerage is based not only on the laws governing 
stress and resistance in solids, but also on the laws governing 
the pressure and flow of water. 

Physical Properties.—Water is one of the fluids classed as 
incompressible because its change of volume under pressure is 
so small that it is neglected in all practical problems. Its den¬ 
sity, however, increases slightly under pressure, and decreases 
slightly as its temperature rises from near the freezing- to the 
boiling-point. 

The weight of a cubic foot of fresh water under normal con¬ 
ditions is 62.425 pounds; in practical problems it is usually 
taken as 62.5 pounds. This is usually an error on the safe side. 

The volume of water is expressed either in units of volume 
or in units of capacity. The unit of volume is the cubic joot in 
all countries where the English system is employed; where the 
metric system is used the unit is the cubic meter. The unit of 
capacity in this country is the standard gallon , which contains 
231 cubic inches; in England it is the imperial gallon , which 
contains 277.27 cubic inches; where the metric system is used 
it is the liter , containing 0.264 standard gallons , or the dekaliter , 
containing 2.64 gallons. 

1 cubic foot = 7.4805 standard gallons = 62.5 pounds. 

1 standard gallon = 0.1337 cubic feet = 8.356 pounds. 

Static Principles.—1. Every molecule oj still water is subjected 
to equal pressure from all directions. 

This results from the hypothesis that in a perfect liquid there 
is no cohesion between its molecules, and that the molecules 
move without the development of friction. This is not strictly 
true since there is slight cohesion between its molecules, but in 
all ordinary problems it may be neglected. 


292 


THE PRESSURE AND FLOIV OF IVATER. 


2 93 


2. The upper or free surface of still water is horizontal or 
normal to the action line of the force of gravity . 

The surface must be normal to the resultant water pressures 
on the surface molecules; otherwise these molecules would 
move along the surface. Since each surface molecule is at rest, 
the resultant water pressure on it must be directly opposed to 
the action line of its weight, which is the only other force acting 
on it. As the action line of the weight is vertical, that of the 
resultant pressure must be vertical and the free surface must 
therefore be horizontal or normal to the action line of the force 
of gravity. 

From this principle it follows that, if the free surface of water 
is not horizontal, the water is not still, but is in motion; if still 
water is subjected to any force whose action line is not vertical, 
the surface will not be horizontal. A flowing river is a body 
of water with an inclined surface, and a pond rippled by the 
wind is a body of water acted upon by force in addition to that 
of gravity. 

3. The unit pressure on each molecule of a body of still water 
varies directly with its depth below the free surface. 

Since there is neither cohesion nor friction, the water pres¬ 
sure on the upper surface of any molecule is due to the weight 
of the molecules above it; as these are of equal weight, the pres¬ 
sure on the molecule must be proportional to its depth. i\ccord- 
ing to the first principle, the pressure from every other direction 
must be equal to this pressure. Thus if 


p = pressure in pounds per square foot, 
p' = pressure per square inch, 
h = depth in feet, 
h ' = depth in inches, 

(1) p = 62.$h, (2) ^ = 777 h, (3) P’ 

1 44 


1728 


h'. 


(503) 


If the weight or pressure of the atmosphere which rests on 
the water be added, the total unit pressure will be p + p 0 or p' + p 0 ', 
in which p 0 and pd are the pressures of the atmosphere in pounds 
per square foot and inch. At the sea-level pd is about 15 pounds. 

The factor h is called the hydrostatic or static head , or simply 




294 


CIVIL ENGINEERING. 


the head; since it varies directly with the pressure, the head 
is often employed in the sense of pressure. 

4. The pressure oj still water is , at every point , normal to the 
surface pressed. 

The pressures on the elementary areas of a plane surface 
will therefore form a system of parallel forces whose resultant 
is a single force. The pressures on the elementary areas of a 
curved surface will in general form a system of non-parallel 
forces which cannot be represented by a single resultant. If, 
however, the curved surface is one of revolution, the total pres¬ 
sure on it may sometimes be represented by a single resultant. 
Thus the total pressure on a semi-cylinder whose axis is vertical 
or whose extreme elements lie in a horizontal plane may be repre¬ 
sented by a single force. 

If the pressure on the semi-cylinder is assumed to be uni¬ 
form, or the same on each unit of area, the total pressure may 
be represented by a single force whatever be the position of the 
cylinder. 

5. The total pressure on any plane surface subjected to the 
pressure of still water is equal to the weight of a prism oj water, 
whose base is the area under pressure, and whose height is the depth oj 
the center of gravity of the area below the free surface oj the water. 

In Fig. 85 let CD be a plane whose 
center of gravity lies in the horizontal 
line E. 

From the third principle we have 
for the pressure of the elementary area 
da of the plane, at a depth h below 
the surface, 

dp = 62. $hda. 

For the total pressure P we have 

P = 62.$Ehda. 

Since I ho a is the sum of the products of each elementary 
area by its distance from the free surface, it will, from the prin¬ 
ciple of the center of mass, be equal to AH, in which A equals 
the area of the plane in feet, and H is the distance in feet of its 
center of gravity below the free surface. Substituting we have 

P = 62.$AH, 












THE PRESSURE AND FLOW OF WATER. 


295 


in which the second member is the weight of a prism whose 
base is A and whose height is H. In the figure H is equal to 
EF. 

Since H is the head at the center of gravity, we may also 
state that the pressure on a submerged plane is the product of its 
area, the head at its center of gravity, and the weight of a unit 
volume of water. 

6. Ij pressure be applied to a unit area 0) the surface oj water 
completely fillmg a closed vessel , an equal pressure will at once 
be felt at every other unit area both within and at the surface of 
the water , and hence on every unit of area of the vessel in contact 
with the water. 

This results from the incompressibility of water and the 
movement of its particles without developing friction. It is 
the principle employed in the hydraulic or hydrostatic press. 

7. The resultant pressure of still water on a plane area acts 
at the center of percussion of he area with respect to the axis formed 
by the line of intersection of the plane and free surface. 

As the elementary pressures form a system of parallel forces, 
they must have a single resultant; the point in which this result¬ 
ant pierces the area is called the center of pressure. 

As the pressure is zero at the surface and increases uniformly 
from that surface, were the area free to rotate under the action 
of the pressure it would rotate about its intersection with the 
surface. This line may therefore be considered its spontaneous 
axis , and the point where the resultant pierces the area, or the 
center of pressure , then becomes the center of percussion. 

The distance of the center of percussion from its spontaneous 
axis may always be found from the following equation derived 
from Mechanics 

/ r* + d'* 

■ D Ad' & ..( S ° 4 ) 

in which D = distance of center of percussion from the spon¬ 
taneous axis, or in this case the distance of the 
center of pressure from the intersection of the 
plane and the water surface; 
d' = distance of center of gravity of area from the 
same line; 




296 


CIVIL ENGINEERING. 


7 = moment of inertia of area about the same line; 

Ad'= moment of area about the same line; 

r = radius of gyration of the plane area about an 
axis through its center of gravity parallel to its 
spontaneous axis. 


As the value of d! for any plane area can easily be determined, 
and as the values of r for all simple areas are given in engineering 
manuals, it is easy to determine the value of D for any simple 
area. 

In Fig. 86 let CD be a rectangular plane area whose edges 
projected in C and D are parallel to the surface AB. 

Let d = length of edge CD, 

b = length of edge C. Then 

A = bd, 

2 d2 
r 2 =—, 

12 

d' = FG, 
d 2 



Fig. 86. 


D = 


12 


+ FG 2 


FG 


If the upper edge of the plane is in the surface AB, FG becomes 
, d , , 

equal to —, and the value of D reduces to 


D = \CG = \d. 


Hence the center of pressure of a rectangular surface whose 
upper edge lies in the free water surface is two-thirds the depth 
of its lower edge below the surface. 

The center of pressure of a curved surface can be accurately 
determined only when the elementary pressures can be replaced 
by a single resultant. 

8 . Archimedes' Principle. Every solid, floating or immersed 
in water, is subjected to a vertical jorce acting upwards, whose 
intensity is equal to the weight of the water displaced by it. This 
is called the buoyant effort of the water. 

When at rest a solid lighter than water will therefore displace 









THE PRESSURE AND FLOW OF WATER. 


297 


its own weight of water. A solid heavier than water will lose 
in weight an amount equal to the weight of an equal volume of 
water. 


PROBLEMS. 

60. What is the static pressure per square inch in sea-water 
at a depth of 50 feet? Weight 64 pounds per cubic foot. 

Ans. 22.2 pounds. 

61. What is the total pressure upon the surface of an inclined 

isosceles triangle whose vertex is at the surface and whose base 
is parallel to and 9 feet below the surface? The base of the triangle 
is 10, and its altitude 12 feet. Ans. 22,500 pounds. 

62. The base of a conical vessel filled with water is a circle 
whose radius is 6 inches; the free surface of the water is a circle 
whose radius is 1 inch. If a pressure of 10 pounds is applied to 
the free surface, what will be the resulting pressure on the base? 

Ans. 360 pounds. 

63. The moment of inertia of a circle about an axis through 

7 rd* 

its center is —. Where is the center of pressure of an inclined 

submerged circular plate whose radius is 10 feet, whose center is 
4 feet below the free surface, and whose circumference is tangent 
to the free surface? Ans. 12.5 feet. 

Discharge through Orifices.—An orifice is an opening in the 
side of a vessel containing water; it is entirely below the free 
surface of the water. The stream which issues from an orifice 
is called a jet. 

Torricelli's Law. The theoretical velocity of the jet is the same 
as that developed by a body falling in vacuo through a height equal 
to the depth of the center of the orifice below the free surface of the 
water. Or 

V = V 2 gH } 


in which V = velocity of discharge in feet per second; 

H = depth in feet of center of orifice below the surface; 
g = acceleration due to force of gravity = 32.2 feet per 
second. 

The theoretical discharge per second is the product of this 
velocity and the area of the orifice; or 




298 


CIVIL ENGINEERING. 


in which D — discharge in cubic feet per second; 

<2 = area of orifice in square feet; 

V = theoretical velocity in feet per second. 

If the vessel is made of thin plates and the orifice is not near 
the bottom or a side, the actual form of the jet is shown in A, 
Fig. 87. The different molecules reach the orifice by converging 


A 

• . Fig. 87. 

paths and make the cross-section of the jet at a short distance 
from the orifice sma er than the orifice itself. 

By comparing the actual path of the jet with its theoretical 
path, it is found that the velocity at the contracted section neces¬ 
sary to produce this path is only 0.97 of the theoretical velocity. 
By careful measurement the area of this contracted section is 
found to be only 0.64 of the area of the orifice itself. The actual 
discharge when measured is found to agree with the actual velocity 
and actual area of the contracted section, and is only 0.62 of 
the theoretical discharge. 

If, therefore, V' = actual velocity at contracted section, 
a'= actual area at contracted section, 

D'= actual discharge, 

V = theoretical velocity at the orifice, 

G = area of orifice, 

D = theoretical discharge, then 

F'=c„F=0.97F, .(505) 

a'=c„a=0.64a,.(506) 

D’=cD=o.62D, . (507) 



£ CyC at 


• (508) 



















THE PRESSURE AND FLOW OF WATER. 299 

in which c v is the coefficient of velocity; 

c a is the coefficient of contraction; 
c is the coefficient of discharge. 

The value of the actual discharge may be expressed as follows: 

D' =cD = caV = ca\ / 2gH, .... (509) 

from which the discharge of any orifice may be computed when 
the coefficient of discharge, the area of the orifice, and the depth 
of the center of the orifice below the water surface are known. 

If the plate in which the orifice is made is thin and clean, 
the edges of the orifice are sharp, and the orifice is not too near 
the bottom or a side of the vessel, the coefficient of discharge 
will be 0.62. 

This value of the coefficient cannot be employed, however, 
if the thickness of the plate exceeds the diameter of the orifice, 
or if the edges of the orifice in a thick plate are not normal to 
the surface of the plate in contact with the water. 

The formula for the discharge of an orifice is usually put in 
the form 

D = 0.62a V2gH = o.62aV, .... (510) 

in which D = actual discharge in cubic feet per second; 
a = area of orifice in square feet; 

H = head in feet over center of orifice; 
g = 32.2 feet per second; 

V = theoretical velocity due to head. 

From this formula the value of any one of the quantities D , 
a, H, or V may be found when the others are given. 

If the orifice is near either the bottom or one of the sides of 
the vessel, as shown in B , Fig. 87, the contraction is only partial , 
and the discharge will be greater than that given by formula 
( 5 10 ). 

Short Tubes.—Short tubes are often inserted into orifices. 
These are of two varieties: the re-entrayit tube , which projects 
into the vessel and has its outer end flush with the outer face of 
the vessel, and the projecting tube, whose inner surface is flush 
with the inner face of the vessel. 




300 


CIVIL ENGINEERING. 


The Borda tube ( C * Fig. 87) is a cylindrical re-entrant tube 
so short that the jet passes through the tube without coming 
into contact with its sides. Its coefficient of discharge is only 
0.5, which is less than that of any other form. This small value 
is due to the crowding of the molecules seeking entrance. The 
formula of discharge for a Borda tube is 


D = o.$a \ / 2gH = o.^ ) aV .( 5 11 ) 

The standard tube is a short projecting cylindrical tube (D, 
Fig. 87). As the jet enters the tube it is contracted as in leaving 
an orifice, but before it leaves the tube it expands so as to com¬ 
pletely fill the cross-section at the outlet. The actual contrac¬ 
tion is therefore nothing, and the coefficient of discharge must 
be equal to the coefficient of velocity. By actual measure¬ 
ment the common coefficient has been found to be o 82. The in¬ 
crease in the discharge over a simple orifice is explained by the 
formation of a vacuum in the tube at the point where the jet 
is contracted; this reduces the back pressure. The vacuum is 
caused by the rushing water carrying with it the confined air. 
The formula for the standard tube is therefore 


D = 0.82a \ / 2gH = 0.82a V .( 5 12 ) 

By changing the form of the projecting tube its discharge 
coefficient may be modified. If made bell-shaped, like the verti¬ 
cal section of a jet when issuing from an orifice ( A , Fig. 87), 
the coefficient will be about 0.95 if the discharge is measured in 
terms of the area of its smallest cross-section. 

Coefficients have been determined, by careful measurement, 
for many forms of projecting tubes, convergent, divergent, and 
combined. These are found in engineering manuals. An orifice 
in a plate whose thickness is greater than the diameter of the 
orifice is simply a projecting tube and has the same coeffi¬ 
cients. 

If the tube is near the top, bottom, or sides of the vessel, the 
molecules will enter the tube with less interference, the coeffi¬ 
cient will be increased, and the jet will be only partially con¬ 
tracted. 




THE PRESSURE AND FLOW OF WATER. 


3 QI 


PROBLEMS. 

64. What will be the discharge in gallons per second from an 
orifice 2 inches square, whose center is 4 feet below the surface? 

Ans. 2.06 gallons. 

65. What head will be required to make the discharge 
through a 2-inch Borda tube one gallon per second? 

Ans. 2.29 feet. 

66. What is the diameter of a standard tube which under 
a head of 9 feet discharges 2 gallons per second? 

Ans. 1.57 inches. 


Miner’s Inch.—The miner’s inch is the discharge in a unit 
of time through an orifice, one square inch in area, under a head 
fixed by law or custom; it is employed in measuring water sold 
for mining and irrigating purposes. A head of six inches is often 
prescribed, which makes the discharge per second 



0.62 

144 



= 0.0244+ cubic feet.. 


(513) 


This formula is deduced under the hypothesis that the coeffi¬ 
cient of discharge is that of an orifice in a thin plate. In prac¬ 
tice the shutter through which the water flows is not a thin plate, 
but a board at least an inch thick. If the coefficient is assumed 
as 0.72, a mean between the coefficients for an orifice in a thin 
plate and that for a standard tube, it will correspond more 
nearly to the conditions of practice. 

The practical method of measuring the discharge of a stream 
in miner’s inches is shown in Fig. 88. The water is discharged 
through an orifice, shown by the shaded area, 
one inch deep whose center is 6 inches below 
the top of the plate, B. A sliding shutter, A , 
in the plate permits of the widening of the 
orifice until the entire discharge takes place 
through it when the surface of the water is 
on a level with the top of the plate, B. p IG< gg # 

The number of miner’s inches is at once 
read off on the scale. 

Weirs.—A weir is a notch cut in the side of a vessel, through 
which the water flows. It is usually rectangular in shape with 


















3 02 


CIVIL ENGINEERING. 


vertical sides; the bottom is called the sill. Triangular notches 

are also employed. 

A weir of the form shown in A , 
Fig. 89, is called a weir without end 
contractions; one constructed as shown 
in B , Fig. 89, is a weir with end con¬ 
tractions. 

Francis’ formula, commonly used for a weir with end con¬ 
tractions, is 

.( 5 i 4 ) 

The end contractions are assumed to have the effect of short¬ 
ening the length of the weir. For weirs without end contractions 
this becomes 

P-3.336.ff*.(515) 

In this formula H is the head of the water in feet, with respect 
to the level of the sill, taken just above the point where the fall 
of surface, due to the weir, is perceptible (Fig. 90); b is the width 
of the weir in feet. H must be measured with great care when 
accuracy is desired. For this purpose a hook gauge is employed. 
This is a graduated rod, provided with a vernier reading to thou¬ 
sandths of a foot, whose index is a hook with point upwards. 
The hook is raised from underneath the surface until the point 
makes a slight pimple on the surface; the zero of the scale is 
set at the level of the sill - 

Weirs are employed to measure the discharge of a stream 
not easily measured by the discharge of an orifice; tables of 
coefficients for different lengths 
are given in the engineering man¬ 
uals. For the accurate measure¬ 
ment of small streams the weir 
is cut in a thin plate of metal, and 
H is carefully measured with a 
hook gauge. For a rough deter¬ 
mination H may be measured as 
shown in Fig. 90, in which h = H. 

The formulas above given are for weirs in the dams of reser¬ 
voirs in which the water is still, and the velocity over the weir is 











































THE PRESSURE AND FLOIV OF IVA TER. 303 

that due to its head above the level of the weir alone. If the 
weir is placed in a running stream, the velocity at the weir for 
the same head over the sill will be greater than in the case of 
still water, and the discharge will be greater than that given 
by formulas (514) and (515). If the velocity of the stream is 
measured, and the height H\ due to that velocity is computed, 
the discharge over the weir is ascertained by substituting (H + H 1) 
for H in formulas (514) and (515) 

PROBLEMS. 

67. What is the discharge in cubic feet per second of a weir 3 
feet long under a head of 6 inches if there is no end contraction? 
If it is contracted at the ends? 

Ans. 3.489 and 3.414 feet per second. 

Flow of Water through Pipes.—In Fig. 91 let the water level 
in the reservoir and the diameter of the pipe both be constant 



and the pipe be running full; assume that there is no resistance 
to the flow of water into the orifice or entrance of the pipe, and 
that there is no resistance in the pipe itself. The velocity of the 
current in the pipe will then be constant throughout the pipe, 
and uniform in each cross-section; its value will be 

V 2 

V 2 = 2gH, and the head which produces it H = —, (516) 

2 S 

in which H equals the head in feet, V equals the velocity in 
feet per second due to the head H , and g equals 32.2. 

Loss of Head at Entrance.—In the above hypothesis the 
error was made of assuming that there was no resistance offered, 
and therefore no pressure required, or head expended, in forcing 
the water into the pipe. 
















3°4 


CIl/IL ENGINEERING. 


The entire pipe BC may be considered as made up of a standard 
tube BE to the end of which is fastened a pipe of equal diameter 
into which the standard tube discharges. In the discussion of 
the standard tube it was shown that the resistance in the tube 
was sufficient to reduce the discharge to 0.82 of the theoretical 
discharge. As the tube runs full at its end E, there is no final 
contraction, and the coefficient of contraction must be equal to 
unity. Since c = c v c 0 , the coefficient of velocity must therefore 
be equal to that of discharge, and the velocity of the water as it 
leaves the tube BE to enter the pipe EC will be only 0.82 of the 
theoretical velocity. As the pipe is of uniform cross-section, 
this velocity will remain constant, and will be the velocity of 
the stream when it leaves the pipe at C. 

Let v = velocity in the pipe or the velocity of discharge; 

V = theoretical velocity; 

h = velocity head; 

h'= entrance head or head lost at entrance; 

V 2 

H = h + h' =—. 


Then for a standard tube 


v = o.&2V, 
^= 0 .6 7 F 2 = §F 2 , 
F 2 = \v 2 , 




3 ^2 
2 2 g’ 


or 


k'=*H-h = ^ — 
2 2 g 

h! = \h. 



1 v 2 

2 2 g* 


Hence we see that the head expended in forcing the water 
into and through the standard tube, which is called the head lost 
at entrance or head due to influx, is about 0.5 of the head required 
to produce the velocity of efflux or discharge. It follows from 



THE PRESSURE AND FLOW OF WATER. 


3°5 

this that if the velocity of discharge is small, the loss of head at 
entrance will also be small, and conversely. 

If the pipe has a bell-shaped or other inlet whose coefficient 

v 

of discharge and velocity is not 0.82 but c } since v = cV, or — = F, 

c 


and H = 


Zr 


C 2 2 g' 


. . . (517) 


for the loss of head at entrance. The expression — d may 
be represented by m. Equation (517) then reduces to 


mv* 

h' = H—h= - =mh .(518) 

In the above discussion, the resistance in the pipe EC has 
been neglected. It is clear, however, that this resistance cannot 
affect the value of m or the ratio of h', the head lost at entrance, 
to h, the head which is required to produce the velocity of discharge. 
This ratio is dependent solely on the coefficient of discharge of 
the inlet-tube. 

Loss of Head Due to Friction.—In the original hypothesis 
a second error was made in assuming that the flow of the water 
met with no resistance in the pipe. As a matter of fact the 
molecules at and near the surface meet with considerable resist¬ 
ance. As the resistance occurs at the surfaces of contact, it is 
usually called frictional resistance. It is found by experiment 
to be governed by the following laws: 

1. The resistance varies directly with the area 0} the surface 
of contact. It is therefore a function of the product of the 
length and perimeter of a pipe running full. 

2. The resistance is independent of the normal pressure on 
the surface of contact. In this respect it differs from friction be¬ 
tween solids. This peculiarity is accounted for by the incom¬ 
pressibility of water and lack of friction between the mole¬ 
cules. 





3°6 


CIVIL ENGINEERING. 


3. The resistance varies with the roughness oj the surface of 
contact. This roughness causes eddies and retards the flow of 
the molecules near the surface. The retardations affect the 
flow of the other layers of molecules inversely as their distance 
from the surface. The velocity is therefore variable in any 
area of cross-section. 

4. The resistance varies directly with some function of the 
mean velocity. It is usually assumed that the resistance varies 

v 2 

with the function —, as experiments indicate that for the velocities 

of practice the resistance varies with the second power of v. 
In pipes of very small cross-section it varies with the first power 
of v. 

Since the velocity of the current is the same at every cross- 
section, the force expended in overcoming the frictional resist¬ 
ance in the pipe must be equal to this resistance. 

If R = frictional resistance, in pounds, of the pipe beyond 
the standard tube, 

1 1/ /1 * 

nd = inner perimeter of cross-section in feet, 

/ = length in feet of pipe whose resistance is R, 

/' = constant depending on roughness of interior surface, 

^ = mean velocity in feet per second of flow at any cross- 
section, 

g = 2)2.2 = acceleration in feet per second due to force of 
gravity, 
we may write 


Tidlf'v 2 

R =- 

2g 


( 519 ) 


If/' = mean unit pressure, in pounds per square foot on the 
cross-section, required to overcome the frictional 
resistance of the pipe in a length /, 
a = area of cross-section of pipe in square feet, 
h" = head in feet required to overcome frictional resistance 
in length /, 

G = 62.5 pounds = weight of a cubic foot of water, 


the force required to overcome the frictional resistance R will be 

p"a = h"Ga. 




THE PRESSURE AND FLOW OF WATER. 


3°7 


Equating the expressions for the force and resistance we have 


h"Ga= 7 cdlj'— } 
2 g 


V’J*. 

a Cr 2g 

r 


( 5 2 °) 


Substituting / for the constant 77 we have 

Cr 


and 

In these expressions 


,, jlv 2 4 fi V 2 
a 2g d ' 2g 

a h" 2g d h" 2g 
' nd l v 2 4 l v 2 ' 


• ( 5 21 ) 

• (S 22 ) 


/ = the coefficient of friction and is a constant whose value 
must be determined by experiment. 

^ = -=ratio of the area of cross-section of interior of pipe to 

its perimeter; this is called the hydraulic mean radius , 
or the mean radius. It is a constant for any given pipe. 
h” 

-j = ratio of head required to overcome the frictional resist¬ 
ance in a pipe to the length of the pipe. It is called 
the hydraulic mean gradient or the hydraulic gradient . 


Many experiments have been made to determine accurately 
the value of the coefficient /. It is found that the theoretical 
expression for it is not sufficiently accurate to admit of the use 
of a single value for all cases. It varies with the material of the 
pipe, its condition of cleanliness, its diameter, and the velocity 
of flow. In each case it is therefore necessary to select from 
the tables found in engineering manuals the empirical value 
which corresponds most nearly to the given conditions. Care 
must be exercised in taking out the proper value, as authors 
employ different expressions for it. 





3°8 


CIVIL ENGINEERING. 


From equation (522) we may deduce the value of /: 

2 g_ dhT 
1 4 v 2 ' l ’ 

t 2 g dh " 

' 2 v 2 l * 

Each of these is employed by some author as the coefficient 
of friction. For their relative values we have 


/ = i/i = 16.1/2 and /1 = 64.4/2. 

For approximate results the following values may be employed: 


f fi f 2 

Smooth pipes.0.006 0.024 0.0004 

Rough pipes.0.012 0.048 0.0008 


These values are somewhat greater than those given in the 
tables. 

Where greater accuracy is desired the following formula 
has been recommended: 


For smooth pipes / = .005 




d = diameter in feet. 


For rough pipes / = .01 


d = diameter in feet. 


Velocity Head.—In the above discussion it has been assumed 
that the water flowed through the pipe at a uniform velocity of v 
feet per second. A fraction of the total head must be expended 
in producing this velocity. 

If i; = the velocity and h = the head, since gravity is the only 
force acting on the water, we have 


v 2 

h =—. 

2 g 


This head h is called the velocity head . 










THE PRESSURE AND FLOW OF WATER 309 

Fanning gives the following table showing the fractions of 
the total head expended in overcoming the various resistances in 
a pipe one foot in diameter when the inlet is a standard tube: 


Length of pipe in feet. 

5 

50 

100 

1000 

10,000 

Velocities, feet per second. . 

63.46 

51.11 

43 -n 

17-38 

5-39 

Head expended at entrance. 

3 i -58 

20.49 

14-57 

2-37 

0.23 

Head expended in overcom- 






ing friction. 

5.88 

38.94 

56.57 

92.94 

99-33 

Velocity head. 

62.54 

40.57 

28.86 

4.69 

0.44 

Total head, H . 

100.00 

100.00 

100.00 

100.00 

100.00 


Formulas.—Since the total head, H , must be equal to the 
entrance head, plus the velocity head, plus the friction head, we 
may write the equation 


1 \v z v^ 
H = h' + h + h" = (- 2 -1 —+ — 

VC 2 /2 g 2 g 


4 jlv 2 
2 dg 


( 4/A v 2 

(0»+i)+ 7 J-, (523) 


or 



(524) 


The coefficient m is equal to 0.5 if the entrance is in the form 
of a standard tube, and 0.08 if in the form of a bell mouth which 
offers little resistance. 

As the standard tube is the common form, it will be substituted 
in the formulas (523) and (524). These formulas then become 





2 g 


) 


( 525 ) 



Since the discharge is equal to the velocity multiplied by 
the area of cross-section, we may also write 







































3 IQ 


C/yiL ENGINEERING. 


D = av = 0.7854 d 2 v = o. 7854 d 2 


2 gH 

_ 4 n 

1 * + t 


= 6.30 3 \- 


5^ + 4//’ (527) 


Solving with respect to d 5 , 


d 5 


1 (i.5^+4//)Z> 2 


(6.303)' 


H 


i • • • 


( 528 ) 


D 2 


d = 0.4788^ (1.5 J + 4//) gr.(5 29) 


To solve this equation a value of d must be assumed for the 
term under the radical sign. The true value of d is obtained 
when its value determined by solving the equation is equal to 
the assumed value. 

Long Pipes.—Fanning’s table, given above, shows that as 
the length of the pipe is increased, more and more of the total 
head H is expended in overcoming the frictional resistance, 
and less and less is expended at the entrance and in giving velocity. 
As the velocity in a city water main is usually only a few feet a 
second, it is customary to consider the head H as entirely expended 
in overcoming the frictional resistance. Hence we may assume 
H = h" and L=/, in pipes having a velocity not exceeding 10 
feet a second, or a length exceeding one thousand diameters. 

Under this hypothesis the four equations become 


4/fo 2 jLv 2 

11 =- T = 0.0021 — 

2 gd d 


H jv 2 

or — = 0.0621—, 
L d 


(530) 


v=yj 


2 gHd 


Hd 


4/Z, 4 - OI2 N 1 ji V 4 ’° 12 \ / 


• 


or 


D =0.7854^ = 0.7854 d 2 
™ l d 5 \H 

D 2 =\g.g2Sjjj-, 


^j2gHd iHd 5 1 


= 3 - 152 ' 


IL 


d- o.4788<P^ 


• o • • © • 


(531) 


(532) 


(533) 




















THE PRESSURE AND FLOW OF WATER,. 


3H 

The practical problems of flow usually appear under one 
of the following forms, in which H, d , and L are in feet, v in feet 


per second, and D in cubic feet 

4 

per second: 

1. 

Given d and — . 

Required 

v and D. 

2. 

‘ ‘ d and v. 

< c 

H 1 ^ 
j- and D. 

3 - 

‘ ‘ d and D. 

< < 

H 

Y and v. 

4 - 

tt H A 

j- and v. 

i i 

d and D. 

5 - 

‘ ‘ — and D. 

i ( 

d and v. 

In forms 1, 4, and 5 both H and L may be given, and in forms 
2 and 3 one of the two may be given and the other required. 


To solve 

I 

employ formulas (531) and (532) 

< c 

< < 

2 

i i 

“ ( 53 °) “ 

( 532 ) 

< < 

< < 

3 

1 < 

“ ( 53 °) “ 

( 532 ) 

(l 

( ( 

4 

< < 

“ ( 53 °) “ 

( 532 ) 

i i 

11 

5 

< < 

“ ( 533 ) “ 

( 53 1 ) 


PROBLEMS. 

67. A pipe 1 foot in diameter has a hydraulic gradient of 
what is the velocity of flow and discharge per second if /= 0.012? 

Ans. 3.66 feet; 2.87 cubic feet. 

68. The velocity of flow in a pipe 2 feet in diameter is 3 feet 

a second; what is its hydraulic gradient and discharge per second 
if/=0.010? Ans. .0028; 9.425 cubic feet. 

69. A pipe 6 inches in diameter delivers 1 cubic foot of water 

per second; what is the velocity of flow and the hydraulic gradient 
if/=0.012? Ans. 5.09 feet; 0.0386. 

70. The hydraulic gradient of a pipe is T io and the velocity 

of flow 4 feet a second; what is its diameter and its discharge per 
second if / =0.010? Ans. 0.994 feet; 3.104 cubic feet. 

71. The hydraulic gradient of a pipe is -pG and its discharge 

is 3 cubic feet per second; what is the velocity of flow and its diameter 
iff=o. OI 2? Ans. 1.017 feet; 3.693 cubic feet. 



312 


CIVIL ENGINEERING. 


Hydraulic Grade Line.—In Fig. 92 let BC be a straight 
horizontal pipe of uniform diameter of which BE is the entrance 
or standard tube. 

Let AB = II; 

<2 

AF = h' + h= (m +1)— ; 

v 2 g’ 


DE = h" 


4 jlv 2 

2 gd ’ 


BC=L ; 

EC=l ; 

■y = mean velocity in any cross-section; 
/ = coefficient of friction of pipe. 


Transposing equation (521) we have 


h” 4 jv 2 
l 2 gd ’ 

h" 

in which -j- is the mean hydraulic gradient, or the tangent of 

the angle made by the hydraulic grade line or virtual slope with 
horizontal line. In Fig. 92, therefore, the line DC is the hydraulic 
grade line. 

Since the head AF is expended in forcing the water into 
the pipe and in giving it a velocity v at E , the remaining head 




at E is DE. Hence if a tube, open at the top, be inserted in 
the top of the pipe at E, the water in that tube will stand at the 
level D. This height or head DE which measures the pressure 
at E in a pipe in which the water is in motion is called the hydraulic 
or pressure head. 

From equation (534) we see that if /, v, and d are constant, 




















THE PRESSURE AND FLOIV OF IVA TER. 


3'3 


the value of h" will vary directly with /; that is, to overcome 
the resistance in any fraction of l will require the expenditure 
of the same fraction of h". Hence we may determine the hydraulic 
or pressure head hi" at any point, as H , from the proportion 

hi" : h" or DE:: HC : CE or 1 


Hence hi" = GH. The loss of head between E and H is there¬ 
fore h" — hi" or DE-GH. 

If a tube, open at the top, be inserted in the top of the pipe 
at H, the level of the water will therefore be at G, the intersection 
of DC and GH. 

As the same is true of every other point of the pipe EC, it 
follows that no water can be delivered at an elevation above DC 
while the water is flowing through BC with a velocity of v. 

Tubes inserted, as above explained, to measure the hydraulic 
or pressure head are called piezometric tubes or piezometers. The 
hydraulic grade line or virtual slope may then be defined as the 
line connecting the water level of a series of piezometric tubes inserted 
in a pipe through which water is flowing. 

If the outlet at C is closed, the water will at once rise in each 
piezometric tube until it reaches the level A ; the water will cease 
to flow in any portion of the pipe BC. The head HI = H at 
any point, as H, is called the hydrostatic or static head. 

If DE is very large as compared with AF, as is the case in 
long pipes, AB and DE may be assumed to coincide and AC may 
be taken as the hydraulic grade line. This is the usual custom 
in practice. 

In equation (534) l is the actual length of the pipe in feet; 
if, therefore, the pipe is either curved, or inclined to the horizon¬ 
tal plane, we must make BC equal to the actual length of the 
pipe. In practice, however, the length of the horizontal projection 
of the pipe between its inlet and outlet is so nearly equal to its true 

AB 

length that the ratio ^7 is not appreciably affected by substituting 

its horizontal projection between inlet and outlet for the true length. 

If equation (534 is solved with respect to v, we have 



2 gd h" 

ITT- 


• • ( 535 ) 




314 


CIVIL ENGINEERING . 


Since the discharge per second is equal to the velocity multi¬ 
plied by the area of cross-section, we see that all pipes of the 
same diameter and the same coefficient of friction will deliver 
the same volume of water provided they have the same hydrau¬ 
lic grade line. If AC be considered the grade line, and the length 
of the pipe does not differ materially from BC, then the velocity 
and the discharge will remain constant whatever be the position 
of the pipe, so long as the outlet is at C, the inlet below A, and 
the whole pipe lies below AC. The pressure in the pipe at any 
point will, however, vary with its depth below the hydraulic grade 
line. 

If the pipe is laid on the line BKC , Fig. 92, it must be treated 
as if made of two pipes united at K. The grade line of the 
part BK is DK or AK , which is higher than that of the pipe BC; 
hence it will deliver water at a higher elevation, but the velocity 
and the discharge will be less. The hydraulic grade line of the 
part KC is higher and steeper than that of BC , and it can deliver 
water at a higher elevation and in greater quantity per second 
than BC, provided it runs full. However, as the pipe BK de¬ 
livers less per second than BC, the part KC will not run full, 
and hence the pressure or hydraulic head in that part of the 
pipe will be zero and the water would not rise in the piezometric 
tubes. If, however, the diameter of the pipe BK be sufficiently 
increased, while that of KC remains the same, the pipe KC will 
eventually run full and the pressure in the pipe will be represented 
by the line KC, and the velocity and discharge will both be greater 
than the velocity and discharge of BC. 

If the pipe is laid along the line AC, the discharge is the 
same as the discharge from BC, but there is no pressure at any 
point of the pipe, and an open conduit could be substituted for 
the pipe. 

The actual pressure in a pipe laid along the hydraulic grade 
line is /> 0 , the pressure of the atmosphere; hence if a closed pipe 
at any point passes above this line, the pressure at that point 
will be less than the atmospheric pressure, air will be disengaged 
from the water and collect at the high point and interrupt the 
continuity of flow. 

Other Losses of Head.—In the discussions above given it 
has been assumed that the level of water in the reservoir is con- 


THE PRESSURE AND FLOW OF WATER. 315 

stant. This is not usually the case. The total head H is there¬ 
fore variable, and in all practical problems it is usually its mini¬ 
mum value which is used. This is the difference of level between 
the inlet and the outlet of the pipe. 

The only loss of head that we have considered in the pipe 
itself is the loss due to friction, and this is the only one that need 
be considered if the pipe is straight and of uniform diameter. 

Ij the pipe has a sharp bend, however, a certain fraction of 
the head is expended in forcing the water around the bend. 
This loss is placed under the usual form 


if 


IT" = n—, 
2g 


(536) 


in which n is a coefficient whose value is determined by experi¬ 
ment. 

Ij the area oj cross-section oj the pipe is changed , the relation 
between the pressure heads and velocities before and after the 
change is given by the equation 

7/i 2 — V-T 

h = ~ — Oh —h\), .( 537 ) 


in which h = loss of total head; 

—-— =loss of velocity head; 

h 2 — h\ =gain of pressure head. 


From this expression h can be determined, when we can measure 
Vi, v 2 , hi, and h 2 . 


For the loss oj 
sudden enlargement 

(537): 


head due to the eddies formed at a section oj 
the following formula has been deduced from 



(538) 


in which h jy =loss of head; 

ai =area of cross-section before enlargement; 
a 9 = “ “ “ after u 


v 2 = velocity after enlargement; 
m" =a coefficient. 






CIVIL ENGINEERING. 


31O 


For the loss oj head due to a sudden contraction a similar 
formula is employed: 



(539) 


in which h v =loss of head; 

a' = area of contracted vein after passing from the 
larger into the smaller pipe; 

02= area of cross-section of smaller pipe; 
c a = coefficient of contraction; 
m'" = a coefficient. 


The value of c a depends on the ratio of the area of cross- 
section of the pipe before and after contraction. 

A fraction of the head is also lost whenever the cross-section 
of the pipe is diminished by the partial closing of a valve. The 
loss due to this cause is also put in the general form and the 
value of the coefficient determined by experiment: 




2 g 


(54°) 


c 


Pipes of Varying Diameters.—In the pipe heretofore con¬ 
sidered the diameter has been constant throughout. In practice, 
however, this is not usually the case; the diameter of a water- 

main decreases as the draft on it 
decreases, and it is therefore made of 
a series of lengths each having its 
own diameter. The lengths are con¬ 
nected by conical pipes called re¬ 
ducers, which make the change in 
the cross-section of the stream at 


-hr 


X 


hr 


X 


Fig. 93. 


each reduction a gradual one, so that discharge in a unit of 
time will be the same at every cross-section. 

To find a pipe of uniform diameter having, under the same 
head, the same discharge, in Fig. 93, let 






















THE PRESSURE AND FLOIV OF IVA TER. 


317 


h, hi hi h ~lengths of sections A, B , C, and D\ 
d\, d 2 , ^ 3 , ^4 = diameters of sections A, B, C, and D; 

V\, v 2 , V3, v 4 = velocities of sections A, B, C, and D; 

hi, h 2} h 3 , /^4 = heads required to overcome friction in A , B y C, 

and D\ 

l y d , v } h = corresponding dimensions of the pipe of uniform 

diameter. 

Under the assumption that the entire head is expended in 
overcoming friction, we have 


h — hi -f- h 2 -f- h 3 + h 4) 


4fl v 2 4hh Vi 2 _ 4 / 2/2 v 2 2 t 4 / 3/3 v 3 2 t 4 / 4/4 v 4 2 

- — 4 — " “ " ■ I ^ - ■ ■ I ■ " * 

d 2g di 2g d 2 2g d 3 2g d 4 2g 

Since the discharge of each section is the same and is 
to the discharge of the uniform pipe, we have 

— d 2 v=—di 2 V\ =—d 2 2 v 2 =—d 3 2 v 3 =-d 4 2 v 4 . . . 

4 4 4 4 4 

Hence 

d 2 d 2 d 2 d 2 

v '= v d? V2=V T 2 ’ V3=V M’ Vi=v dT ■ 


(541) 

equal 


(542) 


( 543 ) 


If in equation (541) we assume that / = /i =/2 = /3 = /4, which 
may ordinarily be done, we have, by dividing each term by 

— and substituting for v\, v 2 , etc., the values above deduced, 

2 £ 


lv 2 hv 2 /d 2 \ 2 l 2 v 2 l d 2 \ 2 , hv 2 (d 2 \ 2 t Uv 2 (d 2 \ 2 / s 

d ~ d x W/ + d 2 W) + d 3 w) + di \diV ’ (544) 


d s 

l =d? h 


d s , d s , d s , 
h d? h + d/ 3 + d/ i ’ 


( 545 ) 


l__h_ h_ 

d 5 ~di 5 + d./ 



(546) 


whence 




















318 


CIVIL ENGINEERING. 


from which either l or d can be deduced if all the other quantities 
in the equation are known. 


PROBLEMS. 

72. A pipe 2000 feet long is made of two equal lengths; one 
has a diameter of 8, and the other of 6 inches. What is the 
discharge per second under a head of 20 feet? /=o.oi. 

Ans. 0.71 cubic feet. 

73. Construct the hydraulic grade line in the above problem. 

Ans. h 1 =3.854, h 2 = 16.145. 

Flow of Water in Open Channels. 

Assume that water is flowing in the channel shown in Fig. 94 
at a uniform velocity and with a constant cross-section; or in 
other terms that its regime is permanent. If we consider a small 



volume of water included between two parallel planes of cross- 
section, it is evident, since the terminal sections are similar 
in all respects, that the pressures will be equal, and hence 
the only force tending to move the prism is the component of 
its weight which acts parallel to the surface. 

The intensity of this component will be 

62.5 alh 
~l ’ 

area of cross-section of stream; 
length of prism; 

sine of angle of slope of surface. 

As the velocity is uniform by hypothesis, this force must 
be equal to the resistance which opposes the flow. This resist- 


in which a — 
l = 
h_ 




















THE PRESSURE AND FLOIV OF IVA TER. 


3 X 9 


ance, as in a pipe, is the frictional resistance, so called, at the 
bed and side slopes of the channel. 

This resistance varies with the area of the surface, its rough¬ 
ness, and with some function of the velocity of the flow. In 
determining the resistance, the free surface of the water is omitted, 
as it is found by experiment that the frictional resistance along 
this surface may be neglected. The area of frictional contact 
is therefore equal to the length of the channel between the cross- 
sections considered, multiplied by the length of the perimeter 
A BCD’, this perimeter is called the wetted perimeter. If the 
same function of the velocity be assumed as in the case of pipes, 
the resistance becomes 


plj'v 2 
*g ’ 

in which p = length of wetted perimeter in feet; 

/ = the length of prism in feet; 

/' = frictional constant; 

discharge in cu. ft. 

■z/ = mean velocity of flow, or area of cross . section in sq . f £ 
g = 32.2 feet per second. 

Equating the moving force and the resistance we have 



plj'v 2 

2 g 


Making /= 


62.5 


we have 


jv 2 ah 

Tg = Jv ' * 


( 547 ) 


(548) 


in which / = coefficient of friction; 

a - = R = hydraulic mean radius; 

P 

h 

- =5 = sine of the slope of the surface. 
I 







3 2 ° 


CIVIL ENGINEERING. 


Hence 



or v=cVRS. . . 


( 549 ) 


This last equation is usually called the Chezy formula and 
is the basis of those since devised. As originally employed 
a constant value was given to c. 

In an elaborate series of experiments undertaken by the 
French engineers Darcy and Bazin about the middle of the 
nineteenth century, it was soon developed that c was not constant, 
but its value was greatly affected by the character of the walls 
of the canal. Also that in a channel in which V S was con¬ 


v 


stant, c did not vary exactly with the ratio 

The formula deduced by Bazin from these experiments was 


V J J 


Vrs, 


( 55 °) 


R 


in which a and /? are constants whose value depends on the 
character of the walls of the canal. For English units these 
values are: 



a 

0 

1. Cement and carefully planed wood. 

2. Smooth ashlar, brick, unplaned wood. 

3. Rubble masonry. 

.000046 
.000058 
.000073 
.000085 

.000122 

.0000045 

.0000133 

.0000600 

.0003500 

.0007000 

4. Earth. 

5. Channels carrying detritus and coarse gravel, 
mountain torrents. 



These experiments were made principally in an artificial 
canal 2 meters wide, 1 meter deep, and about 600 meters long. 

About the same time was made the elaborate study of the 
Mississippi River by Captain Andrew A. Humphreys and 
Lieutenant Henry L. Abbot of the Corps of Engineers, U. S. 
Army. 



















THE PRESSURE AND FLOW OF WATER. 


321 


The formula deduced by these engineers is, in its simplest 
form, 



When the results of the American investigations were com¬ 
pared with the French, it was seen at once that Bazin’s formulas 
could not be applied to a channel of the dimensions of the Missis¬ 
sippi, nor the formulas deduced by Humphreys and Abbot to 
the small channels experimented on by Darcy and Bazin. 

The theory of the flow of water in open chanels was next 
made the subject of careful study by two Swiss engineers, Gan- 
guillet and Kutter, who sought an empirical formula which would 
conform to the data obtained by the American and French 
engineers, as well as that furnished by other experimenters. 

They found that the value of c varied with the following 
general laws: 

1. It increases with R, and most rapidly when R is small. 

2. It increases with the decrease in the roughness of the bottom 
and the walls of the channel. This effect is greatest when the 
velocity is small. 

3. It increases with S if R is less than one meter and if the 
bottom and sides of the channel are smooth. 

4. It decreases as S increases if R exceeds one meter, and 
also in small channels if the surface of the bottom and sides 
is very rough. 

The formula deduced by them and known as Kutter’s formula 
is the one now generally employed. It is based on Bazin’s formula, 
but incorporates the data compiled by Humphreys and Abbot. 


v = 


l m 

a + — Awr 
n o 

/ m\ n 

_ l + \ a+ s)TR^ 


Vrs = I —-— \' / RS=c^RS, ( 552 ) 

# I 


I + 


VR/ 


in which v = mean velocity in feet per second; 

R = hydraulic mean radius or hydraulic mean depth in 

feet; 

5 = sine of slope; 












322 


CIVIL ENGINEERING. 


n — coefficient of roughness of bottom and sides; 
a = constant = 41.66 in English measure and 23 in 
metric system; 

/ = constant= Vone meter = 1.81132 in English measure; 
m =constant = .002807 5 i n English measure and .00155 
in metric system. 

Inserting the numerical values the formula becomes 


v = 


■ ^ , 1.81132 , .0028075 

41.66 +--— + 


n 


S 


( " .002807 n 

I + 14I.66 + --~- l~7=- 

V S ) VR 


\ / RS = cVRS. 


( 553 ) 


The value of the coefficient n, as determined from the results 
of experiments, varies with the character of the bottom and 
sides of the channel as follows: 


Well-planed timber. 0.009 

Cement plaster. 0.010 

Same with one-third sand. 0.011 

Unplaned wood. 0.012 

Ashlar and brick. 0.013 

Canvas. 0.015 

Rubble. 0.017 

Canals in firm gravel. 0.020 

Rivers and canals in perfect order free from stones and weeds 0.025 

Rivers and canals in fair condition. 0.0^0 

Rivers and canals in bad condition. 0.0^5 

Mountain torrents. 0.050 


As the formula is a complicated one, numerous tables and 
graphic diagrams have been computed and inserted in engineer¬ 
ing manuals giving the value of c corresponding to channels of 
different materials, different slopes, and different mean hydraulic 
depths. In all ordinary computations these tables are used. 
(See Trautwine’s “ Engineers’ Pocket Book.”) 

Having determined the velocity, the discharge is ascertained 
from the formula 

D = av .(554) 


in which a = area of cross-section in square feet; 
v = velocity in feet per second. 























THE PRESSURE AND FLOW OF WATER. 323 

Kutter’s formula is applicable to the flow in pipes as well as 
to the flow in open channels. 


PROBLEMS. 

74. What is the velocity of flow in a semi-circular brick conduit 
flowing full when the slope is T fo and the radius of the conduit 
is 5 feet? By Bazin’s formula? By Kutter’s formula? 

Ans. 19.87 feet; 21.31 feet. 

75. What is the discharge per second by each formula? 

Ans. 780.3 cubic feet; 836.85 cubic feet. 

Direct Measurement of Velocity.—The velocity at any point 
in the cross-section of a stream may be determined by direct 
measurement. The three methods ordinarily employed are by 
means of floats , current-meters, or Pitot's tubes. 

The stretch selected for making the measurement is one in 
which the slope of the surface is uniform; the form of cross- 
section is constant, and symmetrical with respect to a vertical 
line through the axis of the stream; and the axis of the stream 
is a right line. 

To measure the velocity by means of floats, two parallel lines 
of cross-section, from one to two hundred feet apart, are marked 
by stretching a cord across the stream, or by placing two poles 
in line, on either bank, at each cross-section. The time which 
is required for each float to pass from the upper to the lower 
cross-section is observed and recorded. 

If the surface velocity is desired, some material is employed 
whose specific gravity is only little less than water. It will 
therefore be almost wholly submerged and offer little surface 
for the action of the wind or air. 

If the velocity at any point below the surface is desired, a 
double float is employed. This consists of a surface float which 
offers little resistance to movement through the water, and a 
submerged float which has a large surface exposed to the action 
of the current. The lower float is in the form of a cylinder or 
globe, and is connected to the surface float by a thin wire. 

If the mean velocity in any vertical plane is desired, the float 
is in the form of a weighted pole which extends from the surface 
almost to the bottom. 


324 


CIVIL ENGINEERING. 


The current-meter has a wheel of some form which is revolved 
by the current; the number of revolutions are automatically 
recorded. The wheel may be attached to a pole and thus lowered 
to any point in the plane of cross-section. The current-meter 
is rated by moving it at uniform rates of speed through still 
water. 

Pitot’s tubes as modified by Darcy consist of two vertical 
glass tubes with horizontal arms at the lower extremities; these 
horizontal arms are fixed at right angles to each other. Each 
horizontal arm is so tapered as to leave only a small inlet for 
the water. In each tube near the bottom is a valve which may 
be closed when it has been lowered to the desired level. In 
using the instrument it is lowered into the water with one of 
the horizontal arms pointing up-stream. The valves are closed 
and the instrument is raised so that the difference of level of 
the water in the two tubes may be read on a vertical scale. The 
velocity at the desired depth is deduced from the formula 

v = 0.84V 2gh, .( 555 ) 

in which h = difference of level of the water in the two arms. 

Velocity Contours.—A velocity contour in any plane of cross- 
section of a stream is a line formed of points whose velocity is 
the same. In a circular pipe flowing under pressure the maximum 
velocity will be at the axis and the velocity contours will be con¬ 
centric circles. In an open channel the point of maximum 
velocity is about the depth below the surface; some of the 
velocity contours will enclose this point. 

Discharge.—If a sufficient number of observations are made 
in any plane of cross-section to plot its velocity contours, the 
discharge may be obtained by the formula 


D = Av+A'v'+ A"v"+ e tc., .... (556) 

in which A, A', A ", etc. = the areas between the successive 

velocity contours; 

v , v', v", etc. = the mean velocities of those areas, 

or approximately the velocities 
midway between the contours. 



THE PRESSURE AND FLOW OF WATER. 


3 2 5 


In gauging a river it is customary to divide the cross-section 
by vertical lines into a number of sections of equal width. Prac¬ 
tically this may be done by attaching tags to the cord which 
is stretched across the stream, by driving poles into the bottom, 
by anchoring floats, or by establishing points in the plane of 
cross-section by intersecting range lines fixed by poles on shore. 
The mean velocity of each section is ascertained by floating a 
pole midway between the points established, by measuring the 
velocity in the same vertical plane midway between the surface 
and bottom, or by measuring the surface velocity in the same 
vertical plane and assuming that the mean velocity is 0.9 the 
surface velocity. 

The area of cross-section of a stream is ascertained by making 
soundings at regular intervals with a pole or line. The depth 
and position of each sounding are plotted on cross-section paper 
to a convenient scale. This facilitates the computation of the 
areas of the sections into which it may be divided. 

The discharge is then obtained by substituting the areas and 
mean velocities in formula (556). 

When only approximate results are desired it is assumed 
that the mean velocity of the entire cross-section of a stream 
is about three-fourths of the maximum surface velocity. 


CHAPTER XVIII. 

TIMBER. 


Timber is wood of a quality and size suitable for engineering 
construction. The abundance and cheapness of timber in the 
United States, its strength, lightness, and durability, and the 
ease with which it may be procured and worked, make it one of 
the most valuable of engineering materials. 

The value of any variety of timber depends upon strength , 
durability , cost , and, for interior work, appearance. All of these 
qualities are more or less dependent upon the physical structure 
of the tree itself. 

Structure.—The timber trees of common use in building con¬ 
struction are exogenous , that is, they increase in size by the addi¬ 
tion of rings of annual growth which are formed immediately 
under the bark. The palms and bamboo are the principal excep¬ 
tion to this rule, and they are used only where other woods are 
too expensive. They lack the strength and stiffness essential in 
large structures. In carpentry the exogenous trees are classified 
as sojt and hard woods, the former designation being applied to 
coniferous trees, and the latter to broad-leaved trees. 

Coniferous Trees.—The typical coniferous trees are the pines , 
which are again divided into soft and hard pines. The term soft 
is applied to white pine, and the term hard pine to all other 
varieties. White or soft pines of large size are found along the 
northern boundaries and along the Atlantic coast of the United 
States, and in Canada. It was formerly the most important 
timber in the United States, but is now so scarce that it is used 
only for finishing. It is white, soft, straight-grained, light in 
weight, and easily worked; it does not warp in seasoning. 

The white and red cedar , the fir, and the spruce are medium 
to large-sized trees whose woods have qualities similar to those 
of soft pine. 


326 


TIMBER. 


327 


•The yellow pine , which is a large tree and the strongest and 
hardest of the hard pines, is found along the Atlantic and Gulf 
coasts. It is dark in color, fine-grained, resinous, and inferior 
only to the best broad-leaved trees in strength and stiffness. It 
may be employed wherever large timbers are required, and is 
also used for interior finishings and floors. The bull-pine of the 
Far West and the loblolly-pine of the South are inferior pines 
of the same class. Larch or tamarack and bastard spruce are 
medium to large-sized trees whose woods have qualities similar to 
those of the inferior hard pines. Cypress and hemlock are also 
coniferous trees used to a limited extent in building construc¬ 
tion. Douglass fir or Oregon pine is one of the principal build¬ 
ing materials on the Pacific coast and is being introduced in the 
East. Its qualities are similar to those of yellow pine. 

If a sector of a full-grown pine-tree is carefully examined, at 
the center will be found a small pith; then a great number of 
concentric rings, varying in width and spacing; and finally an 
envelope of bark. The rings are alternately light and dark 
one light and one dark ring representing a year’s growth; the 
light wood is spring growth and is comparatively soft and weak; 
the dark ring is summer growth and is dense and strong. The 
strength of the wood may therefore be measured by the ratio 
of summer to spring wood in a unit of volume. In a cross- 
section, the rings of summer wood are narrow near the pith and 
near the bark; in longitudinal section the rings decrease in thick¬ 
ness from the ground to the top; hence the strongest timber 
will come from the lower part of the tree, midway between the 
pith and the bark. The average amount of summer wood is 
24 to 40 per cent of the entire tree. 

The sap-wood is a zone of light, weak wood, thirty or more 
rings wide, next to the bark; the outer portion of it is the grow¬ 
ing part of the tree. The heart-wood is the inner and darker 
portion of the sector and has no part in the growth of the tree; 
it is much stronger and denser than the sap-wood. Heart-wood 
results from the gradual change of sap-wood due to the infiltra¬ 
tion of chemical substances from the sap. The proportion of 
heart-wood depends upon the age of the tree, forming about, 
60 per cent of an old, long-leaf pine. 

If the sections of a coniferous tree (Fig. 95) are examined 


3 28 


CIVIL ENGINEERING. 


under a microscope, the wood will be found to consist of a num¬ 
ber of parallel vertical tubes, or cells, of wood fiber, called tracheids , 
t arranged in radial lines. The cells are 

from -?tq- to ■§of an inch in length, and 
from to of these dimensions in 
diameter. They are closed at the ends, 
the end walls being thin in comparison 
with the side walls. In the summer 
wood the cells are flattened and have 
much thicker walls than in the spring 
wood; the increased thickness and flat¬ 
tening account for the greater strength 
of the summer wood. The summer 
wood is, however, somewhat weakened 
by resin-ducts which appear on the 
cross-section in grayish spots. At right 
angles to the tracheids are found groups 
of smaller horizontal tubes, called pith , or medullary , rays; among 
the pith-rays are also found horizontal resin-ducts which are 
smaller than the vertical ones. 

As the coniferous trees are composed almost entirely of the 
vertical cells or fibers with few pith-rays, timber from them is 
uniform in structure and is easily worked and split. 

Broad-leaved Trees. —Oak is the best of this class of trees. 
Although porous and of coarse texture, it is heavy, hard, strong, 
and tough; the sap-wood is whitish, the heart-wood brown 
to reddish brown. It shrinks and checks badly, giving trouble 
in seasoning, but is durable and only slightly attacked by insects. 
Although there are perhaps twenty different varieties of oak, the 
market product is generally classified as live , white , and red 
oak. The live-oak possesses the qualities above given to the 
highest degree, but is not widely distributed, nor can timbers of 
great length be obtained from this variety. The white oaks are 
widely distributed, grow to a large size, and are therefore the 
most important of the oaks. Red oak is inferior to white oak, 
in being of a coarser texture, more porous, more brittle, less 
durable, and more troublesome in seasoning. The trees are also 
widely distributed and grow to a large size. 



Wood of Spruce. 
Fig. 95.* 


* Figures from Bulletin No. 10, on Timber, U. S. Dept. Agriculture. 











TIMBER. 


3 2 9 


The other important broad-leaved trees are the ash, basswood, 
beech, birch, butternut, catalpa, cherry, chestnut, elm, gum, hickory, 
locust, maple, poplar, sycamore, and walnut. As a rule these 
woods are of local importance, or their employment is confined 
to a limited field in engineering structures. 

If a sector of an oak-tree is examined, it will be found to have 
a small pith, rings of annual growth composed of spring and 



summer wood, sap-wood and heart-wood. The spring wood, 
however, contains large pores separated by a porous tissue which 
with smaller pores extend into and sometimes through the ad¬ 
joining summer wood (Fig. 96). The strength of a piece of oak 

may be measured by the ratio of its 
solid to its porous area in any section. 
To the pores are largely due the pat¬ 
terns which are found in oak; these 
patterns are more conspicuous in the 




Board of Oak. Top cross-section; 
right face, radial section: front 
face, tangential section. 

Fig. 97. 


Block of Oak, showing 
medullary or pith 
ray. 

Fig. 98. 


spring than in the summer wood (Fig. 97). In white oak the 
pores are numerous and small; in red oak fewer in number, but 





































33° 


CIVIL ENGINEERING. 


larger. The summer wood, excepting the tissue which runs into 
it from the spring wood, is dark and strong. The dark portion, 
divided into strands by the tissue and by the concentric lines 
of short, thin-walled cells, consists of thick-walled cells, or fibers, 
and is the chief element of strength in oak. 

The pith-rays are prominent, appearing on the cross-section 
as grayish lines with tapering ends; and on the radial section as 
bands or mirrors (Figs. 97 and 98). 

Physical Properties. 

Specific Gravity.—The specific gravity of the wood-fiber itself 
is 1.6, but the specific gravity of a unit of volume of wood ranges 
from 0.3 to 1.3. This range is due to the cellular structure of 
wood and the amount of moisture in the specimen. Woods com¬ 
posed of thin-walled cells have the smallest specific gravity. 
The term water-logged is applied to timber whose cells are filled 
with sufficient moisture to destroy its flotation. 

Grain.—Timber is classified as fine-grained when its rings 
of annual growth are close or it is susceptible of polish; and as 
coarse-grained when its annual rings are far apart or it is diffi¬ 
cult to polish. In most timber the grain is parallel to the axis 
of the tree, but spiral and wavy grains are common; the last 
being well illustrated in curly maple. The surface under the 
bark is more or less irregular, depending upon the tendency of 
the tree to preserve any particular form. This tendency is marked 
in the maple, in which the surface depressions are small and 
numerous, giving the tangent boards or veneering sawed from 
it the beautiful figures which are found in bird's-eye maple. 

Color.—Variations in color in wood may be due to the ratio 
of the summer to the spring wood, to the density of the wood- 
fiber, to incipient decay, or to pigments absorbed in its growth. 
When due to the first or second cause, the darker the color the 
stronger is the wood; when due to the third cause, a dark color 
indicates weakness; when due to the last cause, color is no indi¬ 
cation of strength or weakness, but only of the species to which 
the tree belongs. 

Moisture.—Water in wood occurs in three conditions: it 
forms over 90 per cent of the contents of the living cells; it satu- 


TIMBER. 


33 1 


* rates the walls of all cells; it entirely or partially fills the cavities 
of dead cells, fibers, and pores. The total amount of contained 
water remains nearly constant throughout the year, so that the 
time of felling in itself has but little effect upon the durability 
of timber. The more rapid decay of summer-felled timber is 
due to the fact that in summer the conditions of moisture and 
heat are more favorable to decay, and the wood rots before it 
has had time to season. In winter the cold prevents decay, and 
the timber seasons slowly. 

Shrinkage.—Shrinkage is due to the change in the thickness 
of the walls of the cylindrical cells, or fibers, resulting from a loss 
of moisture. The amount of shrinkage increases with the thick¬ 
ness of the cell-walls. As the side walls of each cell are thicker 
than the end walls, the lateral shrinkage is greater than the longi- 





Efleets of Shrinkage. 
Fig. 99. 




-^ 



Formation of Checks. 

Fig. 100. 


tudinal; and as the walls of the cells of the summer wood are 
thicker than those of the spring wood, summer wood shrinks 
more than spring wood. Lateral shrinkage is somewhat checked 
and longitudinal shrinkage increased by the pith-rays, whose 
direction is perpendicular to that of the main cells of the timber. 
A piece of wood in drying is therefore subjected to severe stresses 
in perpendicular planes which cause cracks or checks. The 















































332 


CIVIL ENGINEERING. 


radial shrinkage will be less and more uniform than the tangential, 
since the summer and spring woods alternate in the former direc¬ 
tion and are continuous in the latter direction. The latter causes 
the radial cracks or permanent checks and the warping of tangen¬ 
tial boards (Fig. 99). The more rapid drying out and shrink¬ 
ing of the ends and sides of timbers cause temporary checks , 
which gradually disappear as the timber becomes thoroughly 
seasoned. The former are called end checks and are very com¬ 
mon. (Fig. 100.) 


Mechanical Properties. 

The value of any variety of wood in building construction 
depends largely upon the following mechanical properties. 

Strength. —From the fibrous structure of wood itself and 
from its growth in annual rings, it is apparent that it will not 
offer the same resistance to the various kinds of straining forces. 
It offers the greatest resistance to tensile, bending, and com¬ 
pressive forces which tend to elongate or crush the fibers. Less 
resistance is offered to a shearing force which acts in a plane 
perpendicular to the fibers or grain, and least of all to a shearing 
force which acts parallel to the fibers or grain. From the com¬ 
plex nature of the structure of wood, a considerable variation in 
strength must be expected in different pieces cut from the same 
tree, even if subjected to the same straining forces. Uniformity 
of strength can be secured only by careful selection. 

The table on page 333 gives the breaking and safe unit work¬ 
ing stresses of oak, white pine or spruce, and yellow pine. 

From experiments on bamboo Professor Johnson draws the 
conclusion that bamboo in its natural form under a bending 
force is twice as strong as oak, weight for weight, when the oak 
is taken in specimens of solid square cross-section. The same 
holds true for crushing strength parallel to the fibers. 

Stiffness under Bending Forces. —A certain degree of stiff¬ 
ness under bending forces is essential in timbers exposed to con¬ 
siderable stress, to prevent the undue deformation of the structure. 
The stiffness of wood ordinarily varies directly with its weight 
and strength to resist flexure, and inversely with the amount of 
moisture in it. 


TIMBER. 


333 


' With grain 


Tension. . .... i 


. Across grain 


Oak 

White pine 
Yellow pine 

Oak 

White pine 
Yellow pine 


' With grain 

Compression. . < 


_ Across grain 


Oak 

White pine 
Yellow pine 

Oak 

White pine 
Yellow pine 


Shearing, 


' With grain 

. . 

_ Across grain 


Oak 

White pine 
Yellow pine 

Oak 

White pine 
Yellow pine 


Bending. 


Torsion 


( Oak 

-j White pine 
( Yellow pine 

f Oak 

•1 White pine 
( Yellow pine 


Modulus of longitudinal elas¬ 
ticity. 


Oak 

White pine 
Yellow pine 


Breaking Unit 
Stress. 

Safe Unit 
Stress. 

Factor of 
Safety. 

10,000 

1,000 

10 

7,000 

700 

10 

12,000 

1,200 

10 

2,000 

200 

10 

5°o 

50 

10 

6oo 

60 

10 

4,500 

900 

5 

3,500 

700 

5 

5,000 

1,000 

5 

2,000 

500 

4 

800 

200 

4 

1,400 

350 

4 

800 

200 

4 

400 

100 

4 

600 

150 

4 

4,000 

1,000 

4 

2,000 

5 00 

4 

5,000 

1,250 

4 

6,000 

1,000 

6 

4,200 

700 

6 

7,200 

1,200 

6 

1,800 

450 

4 



4 




1,200 

3 °° 

4 

900,000 



800,000 



1,200,000 




Hardness.—Hardness measures the ability of the wood to 
resist abrasion and indentation, and is an important quality 
in floors, tenons, mortises, etc. The hardness of wood also 
varies directly with its weight, and inversely with the amount of 
moisture in it. The maximum hardness of any specimen is 
developed by placing the fibers in the direction of the applied 
force. 

Flexibility.—Flexibility is the property of bending without 
rupture. This property of wood is developed by moistening, by 
steaming, and by reducing the material to a form in which one or 
both of the dimensions of the cross-section are small as compared 
with the length. 



























334 


CIVIL ENGINEERING. 


Toughness.—Toughness measures the capacity of the wood 
to resist shock; it is a function of both strength and flexibility. 

‘ Defects. 

Defects in timber are due to peculiarities of growth and 
treatment, to the action of moisture, and to the action of insects. 

Defects due to Growth and Treatment.—These are wind- 
shakes, circular cracks separating the annual rings which are 
caused by the twisting action of the wind; belted timber , a term 
applied to timber which has been killed by girdling before fell¬ 
ing; knots , defects caused by the growth of the wood at limbs 
which impair the ease of working and strength; twists , a term 
applied to timber having a spiral grain; heart-shakes , splits in 
the center of the tree; star-shakes , splits radiating from the 
center; checks , cracks due to improper seasoning; rind-gall, a 
swelling caused by the growth of layers over the place where a 
branch has been removed; waney timber , a term applied to 
boards or beams which are not of uniform cross-section. This 
defect is found in boards cut from the surface of the tree. 

Defects due to Moisture.—The defects due to moisture are 
dry and wet rot. Dry rot is fermentation of the sap and decay 
of the surrounding timber caused by bacteria. Dry rot attacks 
only felled timber which has not been thoroughly seasoned, and 
is encouraged by dampness, poor ventilation, or premature 
painting. Wet rot is the slow oxidation of the wood-fiber under 
the action of air and moisture. It may attack either live or dead 
timber, and is common in felled timber exposed to alternate 
wetness and dryness. Decay is’ promoted by moderate heat 
and checked by cold. 

Defects due to Insects.—Timber immersed in the waters of 
the ocean in the torrid or temperate zones is soon destroyed 
by marine insects. These insects generally belong to one of 
two classes—the teredo navalis or the limnoria terebrans. They 
are found in both warm and cold climates, being more plentiful 
in the former; they will not work in foul or muddy water, or in 
a temperature below 40° F. They confine their operations to 
a zone extending from the bottom to a point somewhat above 
the low-water line. The teredo is first deposited upon the tim- 


TIMBER. 


335 


ber in the shape of an egg from which emerges a small worm. 
This worm enters the timber through a pin-hole, but as it pro¬ 
ceeds into the timber it grows, and the hole is enlarged. Con¬ 
sequently timber attacked by the teredo shows no external evi¬ 
dences of its inroads unless examined closely. The teredo lines 
its hole with calcium carbonate, and therefore prefers a cal¬ 
careous shore. The limnoria resembles a wood-louse; it pre¬ 
fers a silicious shore and soft wood; it does not bore, but destroys 
the wood by eating gradually from the surface inwards. Of 
land insects wood-worms, borer-bees, and ants are the most de¬ 
structive; the latter are very common in tropical and subtropical 
countries. 


Durability and Preservation of Timber. 

If protected from destroying insects, timber may be kept un¬ 
impaired for centuries by placing it in a dry, well-ventilated 
place, or by immersing it completely in fresh or salt water. Tim¬ 
ber placed in a damp or poorly ventilated place, or exposed to 
the weather without protection, will decay rapidly. Since the 
decay is caused primarily by the moisture and albuminous sap 
in the timber, the simplest method of preserving the timber is to 
remove this moisture and then cover the wood with an impervi¬ 
ous coating of paint or other substance. This coating will also 
protect the wood from the attack of insects. If the coating is 
applied before the wood is seasoned, it will confine the moisture 
and dry rot will result. 

Seasoning.—Seasoning timber consists in expelling the moist¬ 
ure by natural or artificial means. The rapidity with which the 
moisture may be safely removed depends upon the size, shape, 
and structure of the timber; the ends of sticks always dry 
much more rapidly than the interior. It is impossible to remove 
all water, and even if it were possible, the most thoroughly sea¬ 
soned timber will at once begin to take up water as soon as it is 
exposed to moisture. 

Natural seasoning consists in piling the timber under cover, 
on skids, with strips between successive layers, so that the air 
may circulate freely and dry out the moisture. The time re¬ 
quired depends upon the kind of wood and its dimensions, vary- 


33 ^ 


CIVIL ENGINEERING. 


ing from one to two years for soft woods, and from one and one- 
half to four years for hard woods. Water-seasoning is the re¬ 
moval of sap by the immersion of the timber in water for about 
two weeks. After removal from the water the wood must be thor¬ 
oughly dried before using. Artificial seasoning consists in exposing 
the timber to a current of hot air in a drying-kiln. The tempera¬ 
ture should not be too high; one-inch pine boards require four 
days at i8o° to 200° F. Oak should first be air-seasoned for three 
to six months, and then exposed to a temperature not exceeding 
150° F.; one-inch boards require ten days. Timber may also 
have its sap removed by steaming; this method prevents check¬ 
ing and cracking, but requires subsequent drying. The bene¬ 
ficial effect of seasoning as affecting strength is twofold: by 
shrinkage, a greater number of fibers per square inch is obtained 
in the plane of the cross-section, and the wood-substance itself 
becomes firmer and stronger. The last is the more important 
consideration. 

Preservative Processes.—Seasoning is a protection against 
decay only when the seasoned timber is kept in a dry, well-venti¬ 
lated place, or its surface is protected by an unbroken, impervi¬ 
ous coating of paint , varnish , or other preservative. 

A paint consists of a base , a vehicle , and a solvent , The 
bases commonly employed are red and brown oxide of iron, red 
and white lead, carbon-black, and graphite. The vehicle is usually 
linseed-oil , raw or boiled, made of flaxseed; as substitutes, cotton¬ 
seed- and fish-oils are employed. The solvent is usually the 
spirits oj turpentine , obtained from the long-leaved pines; as sub¬ 
stitutes benzine and naphtha are used. 

Varnish is made by dissolving gum or resin in oil and 
turpentine or alcohol; it differs from paint in being trans¬ 
parent. 

In engineering practice it is often necessary to employ timber 
under conditions in which it is impossible to protect it by a simple 
coating of paint. Special preservative processes have therefore 
been invented which render the exterior coating of paint unneces¬ 
sary. They depend on one or more of the following principles: 
1 st. Expelling the sap and replacing it by substances of a 
durable nature; 2d. Expelling the sap and filling up the pores 
of the wood with a substance which will prevent the entrance 


TIMBER. 


337 


of moisture; 3d. Saturating the timber with salts of a metallic 
base which will combine with the albuminous matter of the sap 
and make it inert; 4th. Introducing an antiseptic liquid or 
salt which will prevent the growth of bacteria. 

The following are the most important methods in use: 

Bichloride of Mercury.—This process, also called Kyanizing , 
consists in steeping the timber for several days under pressure 
in a solution of bichloride oj mercury [corrosive sublimate ); this 
is one of the strongest antiseptics known, and coagulates the 
fermentable substances. The salt is, however, a virulent poison, 
which is an objection to this process. A plant has been in opera¬ 
tion, however, near Lowell, Mass., for over forty years, and no 
bad effects have been recorded under proper handling. Timber 
treated by this process lasts two or three times as long as unpro¬ 
tected timber. The process possesses the defect, common to 
nearly all, of having the sublimate gradually washed out by the 
action of external moisture. 

Zinc chloride.—This process, also called Burnettizing , consists 
in subjecting the timber to a weak solution of zinc chloride under 
pressure, after first subjecting it to steam under partial vacuum. 
The sap is expelled by vaporization, and its place is taken by 
the chloride solution, which has a strong affinity for wood fiber 
and is antiseptic. The objections to the process are that the 
salt will wash out in service, and that the timber is rendered 
brittle if too large a quantity of chloride is deposited in one place. 
The process has been modified in the United States by the Well- 
house, or zinc-tannin , process, which consists in mixing a small 
quantity of glue with the chloride, and afterwards injecting a 
solution of tannin with a view to closing the pores of the wood 
with a kind of artificial leather. This modification is of so 
comparatively recent origin that its efficacy is as yet a matter of 
speculation. 

These two zinc-chloride processes have been extensively used 
in this and foreign countries, and are without doubt successful 
in greatly increasing the life of timber. It is estimated that 
1,500,000 railroad-ties are yearly treated by the zinc-chloride 
method in the United States alone. 

Sulphate of Copper.—This process, also called Boucherie’s, 
consists in forcing a solution of copper sulphate (blue vitriol) 


33 8 


CIVIL ENGINEERING. 


into the timber by hydraulic pressure. This solution drives out 
and replaces the sap. This process has been extensively em¬ 
ployed in France and Germany. Its disadvantages are that 
the salt dissolves out, and also that an iron spike or nail is liable 
to decompose the sulphate, forming sulphuric acid, which attacks 
the wood. The durability of wood is, however, about doubled 
by this process. 

A modification of this process known as the Thilmany process 
consists in treating the timber to a second bath, of barium chloride, 
forming barium sulphate, an insoluble salt, in the ends of the 
pores. 

Creosote.—This process, in one of its forms (Bethel's, Seeley's , 
Hayjord’s , or Creo-resinate ), is in many respects the best process 
for preserving timber, and consists in impregnating the timber 
with heavy oil of tar, known as creosote, or dead-oil. In Bethel's 
process, the earliest used, seasoned timber was first subjected to 
a partial vacuum, and the oil was then introduced at a tempera¬ 
ture of 120°, and under pressure. Seeley's process, designed to 
permit the use of green timber, was a modification in which 
the timber was boiled in dead-oil for a sufficient time to expel 
any moisture; the hot oil was then replaced quickly by cold oil, 
forming a partial vacuum in the cells. The oil was forced into 
the wood by the difference between the external and internal 
pressure, by capillary action, and by the application of extraneous 
pressure to the liquid oil. 

The Hayjord process is a further modification designed to 
prevent timber from checking, which it is liable to do if subjected 
while green to a heat of 212° F. The timber is heated in dry air 
under pressure, and then at a lower temperature subjected to 
a partial vacuum which causes the sap vapors to be withdrawn. 
The dead-oil is then introduced under pressure. 

The creosoting methods have been adopted with a view to 
preventing the loss of the preservative by exposure to moisture; 
the oil, being insoluble, better resists leaching out than the soluble 
salts. So long as the timber retains the oil it will not decay, but 
the creosote will wash out in time, though more slowly than the 
metallic salts. While giving the most satisfactory results, the 
cost of the most approved method (Hayford’s) renders a more 
economical method desirable. 


TIMBER. 


339 


To render the creosote in the timber better able to resist the 
action of the water, experiments are now being carried on at 
various places with a combination of zinc chloride and creosote. 
A further modification of Hayford’s process, known as the creo- 
resinate process , is also being employed, in which, instead of 
creosote, a mixture of creosote (30%) and pulverized resin (70%) 
is used; this, it is claimed, seals the pores with an insoluble water¬ 
proof compound. 

Protection against Limnoria and Teredo.—For subaqueous 
work the creosote method possesses the great advantage of 
protecting the timber against the attack of these insects, so long 
as it remains thoroughly impregnated. The creosoting should, 
however, be thorough; no less than from 15 to 19 pounds of 
oil per cubic foot of wood should be injected for complete pro¬ 
tection. The form of protection, commonly employed for piles, 
is a coating of thin sheets of metal, such as yellow metal, copper 
zinc, etc.; to prevent galvanic action these are secured by nails 
of the same metal. These sheets extend from about the line of 
extreme high tide to 4 feet below the mud line, and are usually 
put on over a layer of tarred felt, which increases the efficacy 
of the protection, and will protect the wood for a time after the 
metal covering is broken. The best results require that the 
sap-wood of the pile be removed. Protection may also be 
obtained by covering with terra-cotta pipe filled with concrete; 
by covering with layers of canvas saturated with some preserva¬ 
tive; or by studding thickly with broad-headed nails. Palmetto 
is the only wood of the United States which in its natural state 
is immune from the attack of the teredo and limnoria. Some 
Philippine woods are said to possess this property. 

Market Forms and Designations. 

Sawing.—Two general methods of sawing are employed- in 
the preparation of timber for the market— bastard , and quarter 
or rift sawing. In the former method the cuts are all parallel to 
each other; in the latter the log is first cut into quarters and then 
each quarter divided by cuts parallel to its bisecting radius, or 
parallel to its faces alternately. The sections made by bastard 
sawing, which is the more common and cheaper method, are 


340 


CIVIL ENGINEERING . 


tangential sections; those made by quarter sawing are approxi¬ 
mately radial sections. Quarter-sawed lumber wears better, 
warps, splinters, and shrinks less, and in most hard woods pro¬ 
duces a handsomer surface. 

Products.— Ceiling; standard sizes are f to f inch thick 
and 4 to 6 inches wide, including tongue. It is matched, and 
dressed on both sides. Flooring; standard sizes are from i 
inch to ij inches thick and 4 to 6 inches wide, including the 
tongue. Stepping embraces all sizes from 1 inch to 2 inches 
thick and from 7 inches up in width; plank , all sizes from 
ij to 5 inches in thickness and from 7 inches up in width; 
dimension timber, all sizes from 6 inches up in thickness and 
7 inches up in width. Timber is generally measured by board 
measure (B. M.), that is, by the number of superficial feet the 
piece would contain if it were sawed into boards one inch thick. 
Thicknesses less than one inch are either counted as one inch, 
or sold by the square foot. 

Specifications.—Specifications vary with the locality, the 
lumber-manufacturers’ associations usually having certain local 
standards. As a rule, each kind of lumber is divided into two or 
three classes according to the allowable imperfections. The terms 
clear, prime, and merchantable are employed to designate the three 
classes in order of value. The symbols S 1 S, 2E, or S 2 S, etc., 
are employed to designate that boards are to be surfaced one 
side and two edges, or surfaced two sides, etc. Surfacing means 
planing by hand or by machinery. In making specifications for 
lumber it is usually best to make them conform to the classifica¬ 
tion established by the lumber association which furnishes the 
lumber for the locality. This prevents unnecessary disputes. 

Tests.—The general character of the material is determined 
by careful inspection. The tests for seasoning consist in care¬ 
fully drying the wood in an oven; the difference in weight before 
and after drying gives the amount of moisture driven off. In 
thoroughly seasoned lumber this should not exceed 10 per cent. 
Tests for crushing, bending, shearing, and tensile stresses when 
necessary are made with testing-machines. 


TIMBER. 


341 


Joints and Pieces in Timber Construction. 

Joints.—Rankine gives the following rules for the construc¬ 
tion of the joints of a wooden frame or truss when its parts are 
subjected to the action of external forces: 

I. Cut the joint and arrange the fastenings so as to weaken 
as little as possible the pieces they connect. 

II. Place the abutting surfaces of a compression-joint as 
nearly as possible perpendicular to the action line of the pressure 
which it must transmit. 

III. Proportion the area of the joint to the pressure which 
it must resist, and form and fit the surfaces accurately so that 
the stress is distributed uniformly. 

IV. Proportion the fastenings of a tensile joint so that they 
may be as strong as the pieces they connect. 

V. Arrange the fastenings so that the joint shall not give 
way by the fastenings shearing or crushing the pieces. 

To this may be added Tredgold’s caution: 

VI. Joints should be so formed that the expansion or con¬ 
traction of timber will not injure the pieces connected. 

Lap-joint.—A lap-joint (Fig. 101) is made by overlapping 
the pieces and fastening them together by bolts or straps. It is 
a simple, strong joint for rough work. In military bridges a 










£-£ 

V, 

TTTTTT7TT 1 


Fig. ioi. 


lap-joint is often made with rope-lashings instead of bolts and 
straps. 

Butt-joint.—A butt-joint (Fig. 102) is one in which the ends 
of the pieces are cut square and abut against each other. The 



Fig. 102. 






I' 

W3— H-r 

I' I 

-C3- 


Fig. 103. 


pieces are held together by fish-plates which overlap the joint 
and are bolted together through the pieces. The fish-plates 









































342 


CIVIL ENGINEERING. 


may also be notched into the pieces, or have keys notched into 
the fish-plates and pieces, as in Fig. 103. 

In order that a tensile joint like Fig. 102 shall be equally 
strong in all its parts, the tensile strength of the pieces, the 
tensile strength of the two fish-plates, the shearing strength of 
the bolts, and the bearing value of the bolts must all be 
equal. If keys are employed as in Fig. 103, the stress on the 
bolts, and the strength of the fish-plates and pieces, are all 
reduced. 

Scarf-joint.—A scarf-joint (Fig. 104) is a lap-joint in which 
the overlapping parts are so reduced that their combined area 
at the joint is no greater than the area of cross-section of either 
piece. The scarf-joint may be strengthened by bolts, straps, or 
fish-plates, and the pieces may be notched into each other. A 


1— ^ — 

?■- ‘r- -^1 

1 !! 

u 



- p - 7 

% 

! 11 


!! 1 

»— i — 

-e 

=—dL— 

3 --* 


Fig. 104. 


simple scarf-joint is one in which the plane of contact of the two 
pieces is a single inclined plane. The opening of this joint, due 
to the shrinkage of the pieces, is almost imperceptible. 

Compression-joint.—A butt-joint, and a scarf-joint like Fig. 
104, are suitable joints for resisting a compressive force if strength¬ 
ened to resist bending by placing fish-plates on all sides. A 
lap-joint may be employed when the compressive force is not 
great. 

Tensile Joint.—A lap-joint and a butt- or scarf-joint properly 
strengthened by fish-plates are suitable joints to resist a tensile 



Fig. 105. 


force. If the tensile force is small, a joint like Fig. 105 without 
fish-plates may be employed. The joint is kept closed by the 
use of folding wedges. 

Flexure-joint.—A joint to resist flexure must have plane 
abutting surfaces, normal to the fibers, in the part of the beam 
subjected to compression, and a fish-plate to transmit the tensile 




















TIMBER. 


343 


stress in the fibers on the opposite side of the neutral axis. A 
butt-joint with a fish-plate bolted to the surface whose fibers are 
subjected to the greatest tensile stress, and a scarf-joint in which 
the joint above the neutral axis is a simple butt-joint and the 
lower part is fished as above described, are suitable joints to 
resist flexure. A scarf-joint of the form shown in Fig. 104 has 



Fig. 106. 


been found to be stronger when the fish-plates are parallel to the 
bending force, as in Fig. 106, than it is when the fish-plates are 
perpendicular to the force, as in Fig. 104. In temporary struc¬ 
tures, a lap-joint may be employed to resist flexure. 

Common Mortise-and-tenon Joint. — A mortise is a hole made 
in one piece to receive the tongue , or tenon , which terminates the 
other (Fig. 107). The end surfaces of the piece on either side 




Fig. 107. 



of the tenon are the shoulders; the surfaces adjacent to the mor¬ 
tise upon which the shoulders rest are the cheeks. A pin may 
be employed to hold the tenon in place. The width of tenon 
is usually about one-third of the width of the piece of which it 
forms part; its length is slightly less than the depth of the mortise, 
unless the mortise passes entirely through the beam. 

If the pieces connected by a mortise-and-tenon joint are 
beams subjected to forces of flexure, a modified form of the simple 
joint is employed (Fig. 108). This is called a tusk-tenon; the mor¬ 
tise may extend wholly through the piece as shown, or it may 
extend only part way through. The area of cross-section of the 









































344 


CIVIL ENGINEERING. 


tenon at the shoulders is increased so as to strengthen the beam, 
of which it forms part, to resist the shear at its extremity; the 
axis of the mortise lies in the neutral surface of the beam from 
which it is cut so as to reduce as little as possible the strength 
of this beam to resist flexure. A joint of this description is 
employed in connecting heavy floor-beams. 

Oblique Mortise-and-tenon Joint.—The oblique mortise-and- 
tenon joint is shown in Fig. 109. According to Tredgold the 
depth of the mortise should be equal to one-half the depth of 
the piece in which the mortise is cut; in practice it is usually 
less than this, since a deep mortise greatly reduces the strength 
of the beam. The effect of the thrust in the strut A is to shear 
off the beam B in a horizontal plane and in the vertical planes 




of the sides of the tenon; hence the shearing resistance of these 
surfaces must be greater than the horizontal component of the 
compression in the strut A. If this condition cannot be ful¬ 
filled, the joint must be strengthened by means of a strap around, 
or bolt through, both pieces. The tenon is then omitted. When 
several pieces intersect at a common point, as at the head of a 
king-post, the strap may be designed to unite all the pieces, as 
shown in Fig. no. 

Bridle-pieces.—A tie may be made of two pieces which are 
bolted together and inclose the pieces connected by them; the 
ties are then called bridle-pieces. Blocks are placed between 
the bridle-pieces at the points where the bolts are inserted (Fig. 
iii). 

Halving and Notching.—These are the means employed to 
reduce the joint depth of two beams at the points of intersection. 
This reduces the strength of the pieces cut. The beams form- 
























TIMBER . 


345 


ing the sills of a building are joined by halving (Fig. 112). The 
ends of roof-rafters are often notched (Fig. 113). 



v 'Y\J v ' 


V\A 


\W| 













6 


1 





T 


0 


i I 






Fig. hi. 



If the span is so great that the cross-section of a single beam 
is insufficient to support its load, the beam may be strength¬ 



ened by the use of shores , corbels , or straining-beams (Fig. 114). 

Shores.—A shore is a prop or strut which supports some 
point of a beam between its extreme supports. By this means 
the maximum bending moment in the beam may be greatly 
reduced. Shores are employed when a suitable surface can 
be found underneath the beam upon which the shores may 
rest. 

Corbels.—A corbel is a short beam placed between the beam 
and the top of the pillar or other support. The effect of a corbel 
is to increase the width of the supported portion of the beam 
and thus decrease the maximum bending moment in the entire 






















































346 


CIVIL ENGINEERING. 


beam between the pillars. When the corbels are long, their 
ends are in turn supported by shores. 

Straining-beams.—A straining-beam is one which supports 
the middle part of the beam itself and is in turn supported by 
shores. Not only is the bending moment at the middle point 
of the beam thus diminished, but the resistance of the straining- 
beam is added to that of the main beam thus strengthened. 

Built-up Beams.— Built-up beams are made by superposing 
simple beams and so uniting them that they cannot slide on 
their surfaces of contact. The beams may be united by bolts, 
or the beams may be held together by bands, and the sliding 
prevented, by notching the beams or by introducing keys or 
folding wedges along the plane of contact (Fig. 115). 


_rqp_ 

ji 

_nqp_ 

|i 

! 1 r77771 

-cp- 

-qp- 

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_ qp _ 


11 

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l! 

1 • 


1 1 

hr 1 

V 






If each layer is made of two or more pieces, those in the 
upper layers are united by simple butt-joints and those in the 
lower layers by butt- or scarf-joints strengthened by fish-plates. 

Curved Beams.—A curved beam may be made of several 
layers of boards, thin enough to be bent over a form, which are 
spiked or bolted to each other; or it may be made of beams 
whose axes are chords of the arc, and which are united by in¬ 
clined butt- or miter-joints. Curved beams are employed in the 
construction of military bridges; their ends are joined by suit¬ 
able tie-rods. 

Straps, Bolts, and Shoes.—Joints in timber-work are often 
strengthened by the use of iron straps , bolts, and shoes. Straps 
are made of wrought iron and are in the form of the letter U with 
square angles. The ends are provided with screw-threads and 
are connected by an iron plate held in place by nuts. Bolts 
have been heretofore described. Rectangular plates or washers 
are employed to prevent straps, bolt-heads, and nuts from cut¬ 
ting into the wood. Cast-iron shoes are also employed instead 
of straps at the panel points of wooden trusses. The struts fit 
in sockets formed in the shoes, and the ties are attached to bolts 
which pass through the shoes. 










TIMBER. 


347 


Strengthened Beams.—A beam may be strengthened by 
trussing it as shown in Fig. 116. The stress in the beam is then 
wholly compression and is uniform throughout. In long beams, 



two stanchions or struts, counterbraced, are employed instead 
of one. If the tie is a single rod, it lies in the plane of the axis 
of the beam; if the tie is composed of two rods, they are on 
opposite sides of the beams. 

For further information consult Tredgold’s “Carpentry”; Kidder’s 
“ Building Construction and Superintendence, Part II ” ; Reports of Forestry 
Division, Department of Agriculture, U. S.; Snow’s “ Principal Species of 
Wood ” ; Johnson’s “ Materials of Construction ”; Thurston’s “ Materials of 
Engineering, Part I.” 





CHAPTER XIX. 


METALS. 

Because of their strength and durability the useful metals 
and their alloys form a very important class of engineering mate¬ 
rials. They may be safely exposed to conditions in which timber 
would fail. The principal metallic substances employed in 
engineering are iron , steel , copper , lead , tin, and zinc. 


Iron and Steel. 

The simplest classification of the three varieties of iron derived 
from the ores, based on the percentage of carbon and on their 
distinguishing properties, is given in the following table: 



Percentage of Carbon. 

Properties. 

Cast iron. 

5 to 2 

2 .O tO O . IO 

Fusible, not malleable. 
Fusible to malleable. 

Stppl . 

Wrought iron. 

0.2 tO 0.05 

Malleable, not fusible. 


The lines of demarcation are not sharp; cast iron and hard 
steel on the one hand, and wrought iron and soft steel on the 
other, are somewhat similar in composition and in qualities. 
Besides iron and carbon, the different members of the group 
contain silica, sulphur, phosphorus, and manganese in various 
proportions, which greatly modify the range of strength and 
the qualities of the metal of each group. Taken as a whole, the 
group forms the most important structural material in engineer¬ 
ing practice, because of its strength, its durability, its compara¬ 
tive cheapness, and the ease with which it may be fashioned 
into the forms best suited to resist all classes of straining forces. 

348 












METALS. 


349 


Cast Iron. 

Cast iron is made by remelting, in a cupola furnace, certain 
grades of pig iron, called foundry pig, which are the product of 
the blast-furnace. From the cupola furnace the molten metal 
is poured into cavities formed in molding sand by the use of 
wooden patterns. The products of the molds are called castings. 
If the process of remelting is repeated two or three times, the 
product is improved at each remelting. Scrap-iron is therefore 
usually mixed with the foundry pig. On account of the ease of 
manufacture, cast iron is largely used in engineering practice, 
in the manufacture of hollow columns, pipes, and ornamental 
forms not easily made of non-fusible metal and not requiring 
the strength or toughness of steel or wrought iron. 

Commercially, cast iron is classed as gray and white ; the gray 
is the class ordinarily employed in engineering practice. It is 
soft, tough, and slightly malleable when cold, and may be turned, 
drilled, and planed. It has a gray, granular fracture, with a 
metallic luster. It is usually subdivided into several minor classes 
differing in hardness. White iron is hard, brittle, and sonorous; 
it cannot be worked and has a white crystalline fracture with 
a vitreous luster. It is used only where a very hard metal is 
required. Chilled cast iron is gray iron with a white iron face. 

PROPERTIES. 

Specific Gravity.—The specific gravity varies according to 
the composition and method of casting from 6.9 to 7.5; in ordi¬ 
nary calculations a cubic foot is assumed to weigh 450 pounds. 

Expansion and Contraction.—Under the influence of changes 
of temperature a cast-iron bar will elongate or contract about 
0.0000062 of its length for each degree of Fahrenheit. Above 
120 0 F. it begins to lose strength; at a red heat this loss is 33 per 
cent; below the freezing-point its strength is unreliable. The 
elongation or contraction of a bar of one inch cross-section is 
of its length for each ton of applied force. 

Strength.—The ultimate unit strength of any casting de¬ 
pends on the composition of the metal, the size and shape of the 
casting, and the care exercised in its manufacture. 



35 ° 


CIVIL ENGINEERING. 


The following values are considered safe values to use in engi¬ 
neering practice: 



Breaking Unit 
Stress, Pounds 
per Square Inch. 

Allowable Unit 
Stress, Pounds 
per Square Inch. 

Tension. 

18,000 

80,000 

20,000 

36,000 

25,000 

6,000 

15,000,000 

3,000 

10,000 

3,000 

6,000 

4,000 

Compression. 

Shearing. 

Bending. 

Torsion. 

Elastic limit in tension. 

Modulus of longitudinal elasticity. 



DEFECTS. 

The defects in castings arise from improper composition of 
the metal and careless manipulation in the process of casting. 
The former causes weakness and brittleness, and the latter 
blow- or air-holes , honeycomb or cavities , cracks , flaws, and cold- 
shut. Air-holes, honeycomb or cavities result from air or im¬ 
purities being retained in the molten metal while cooling; cracks 
and flaws result from unequal cooling. Cold-shut is a surface 
of weakness caused by filling the mold with partially cooled 
metal, through two or more apertures; the metal does not cohere 
perfectly at the surfaces of contact. 

DURABILITY AND PRESERVATION. 

Cast iron is durable for an indefinite time if kept in a dry 
place or immersed in fresh water; if exposed to the alternate 
action of the air and moisture, or to the ordinary gases arising 
from the combustion of coal, it rusts; if the gray variety of cast 
iron is exposed to the action of sea-water, it gradually softens 
and becomes porous. It may be protected against the action 
of moisture, gases, and sea-water by coating it with an impervi¬ 
ous layer of preservative. The first coat should be applied soon 
after the casting is made and before it has time to rust; in sub¬ 
sequent applications all rust must be removed before the pre¬ 
servative is applied. Any of the ordinary paints may be em¬ 
ployed for this purpose; water-pipes are usually heated, and 
plunged into a bath of hot coal-tar pitch varnish. White iron 
















METALS. 


35 r 


resists the action of sea-water better than the gray variety. Al¬ 
though durable, cast iron will in time lose part of its strength 
under the action of excessive loads, or when subject to shocks. 

SPECIFICATIONS AND TESTS. 

In Cooper’s “ Highway Bridges ” the specifications for cast 
iron are as follows: 

“All castings must be tough, gray iron, free from cold-shuts 
or injurious blow-holes, true to form and thickness, and of a work¬ 
manlike finish. Sample pieces one inch square, cast from the 
same heat of metal in sand molds, shall be capable of sustain¬ 
ing, on a clear span of 12 inches, a central load of 2400 pounds 
when tested in the rough bar. A blow from a hammer shall 
produce an indentation on a rectangular edge of the casting 
without flaking the metal.” 

This transverse strength is based on a maximum fiber stress 
in the test-bar of 43,200 pounds. Johnson gives the following 
specifications as employed by the St. Louis Water Department: 

“All of the iron castings shall be made from a superior quality 
of iron, remelted in the cupola or air furnace, tough, and of an 
even grain, and shall possess a tensile strength of not less than 
18,coo pounds per square inch. 

“Test-bars of the metal 3 inches by \ inch when broken 
transversely, 18 inches between supports, and loaded in the 
center, shall have a breaking load of not less than 1000 pounds, 
and shall have a total deflection of not less than of an inch 
before breaking.* Said bars to be cast as nearly as possible 
to the above dimensions without finishing; but correction will 
be given by the Water Department for variations in thickness 
and width, and the corrected result must conform to the above 
requirements.” 

The transverse strength is based on a maximum fiber stress 
of 36,000 pounds. 

Imperfections in the metal are discovered by very careful 
inspection and by the tests mentioned. In testing water-pipes, 

* The tensile strength may be raised to 20,000 or even 25,000 pounds per 
square inch, while the deflection may be made f of an inch for ordinary good 
cast iron, and \ inch for a better quality. For a superior quality it may be 
made § inch, with a breaking load of 1250 pounds. 





35 2 


CIVIL ENGINEERING. 


each section is submitted to a hydraulic pressure of 200 to 300 
pounds per square inch, and while under pressure is sharply 
struck at different points with a hammer. 

Malleable Cast Iron—Malleable cast iron is made by a process 
of annealing or partially decarbonizing white iron. By this 
process its toughness and tenacity are greatly increased. 

Steel. 

Structural steel is made by one of two processes, the open- 
hearth and the Bessemer. 

In the open-hearth process certain grades of pig iron are 
melted in a reverberatory furnace, either basic- or acid-lined, and 
either scraps of wrought iron or steel, or pure oxides of iron 
(ores) are added to secure the proper composition. The jrhdLten 
metal is then molded into ingots , which are rolled or jorged into 
the required shapes. In the Bessemer process certain grades 
of pig iron, melted in the cupola furnace, or direct from the blast¬ 
furnace, are poured into a converter, which has a basic or acid 
lining, and there subjected to a blast of air and the addition of 
molten pig or spiegeleisen to secure the proper composition; 
the molten metal is then molded into ingots. The ingots are 
reduced to structural forms by reheating and rolling. The 
open-hearth process is under more perfect control and admits of 
more frequent analyses; its product is therefore more reliable, 
uniform, and homogeneous. For these reasons it is preferred by 
engineers for structural work. Bessemer steel is largely used for 
steel railway rails. 

Steel wire is made by drawing thin rods through slightly 
conical holes in steel plates; the diameters of the rods are gradu¬ 
ally reduced by successive drawings. The tensile strength of 
the material is greatly increased by the process. 

Structural Steel Divisions.—Structural steel is usually divided 
into rivet, low or sojt, and medium steel, depending upon its 
tensile strength and ductility. 


Kind. 

Tensile Strength. 

Elongation. 

Rivet or very soft. 

Soft, low, or mild. 

Medium. 

48,000 to 59,000 lbs. 
54,000 to 62,000 “ 
60,000 to 70,000 * ‘ 

26 per cent 
25 “ “ 

22 “ " 

















METALS. 


353 


These limits are not strictly adhered to either by engineers or 
manufacturers. Steel having a greater tensile strength and less 
elongation is called high or hard steel. 

PROPERTIES. 

Specific Gravity.—The specific gravity of steel is about 7.8; 
in ordinary calculations a cubic foot is assumed to weigh 490 
pounds. 

Expansion and Contraction.—The expansion or contraction 
of a bar per unit of length for each degree Fahrenheit of change 
of temperature is 0.0000067; the expansion and contraction of 
a bar of one square inch cross-section is about t3-ttotf f° r each 
ton of applied weight. 

Ductility.—Ductility is an important property of structural 
steel, since, combined with strength, it gives the toughness which 
is essential in all structures subject to shocks and moving loads. 
The percentage of elongation under a tensile stress which will 
be required of the steel is always stated in the specifications. 
This percentage increases as the steel approaches the composi¬ 
tion of wrought iron, or as the amount of carbon is decreased. 

Hardness and Fusibility.—The hardness and fusibility both 
increase with the percentage of carbon, or as the metal approaches 
the composition of cast iron. 

Hardening, Tempering, and Annealing.—Steel is hardened by 
suddenly cooling it from a temperature of about 1300° F., called 
a low yellow heat. For this purpose it is plunged into a bath 
of mercury, water, or oil. It may then be annealed or have its 
original properties restored by reheating it to a low yellow heat 
and cooling it slowly; for the latter purpose, when removed 
from the furnace it is covered with quicklime or charcoal. The 
hardened steel may also be tempered by reheating to a tempera¬ 
ture below 700° F., called a dull red heat , and cooling it slowly. 
The amount of softening depends on the range through which 

it is cooled. 

Weldability.—Soft steel has the property of weldability, that 
is, two pieces being heated and superposed may be united by 
repeated blows of a hammer. Medium steel can be welded 
only imperfectly by ordinary processes, and hard steel not at all. 



354 


CIVIL ENGINEERING. 


Electric welding, which consists in melting the surfaces by an 
electric current and uniting them by pressure, can be applied to 
medium and hard steel. welded joint can be relied on to 
carry one-half the load of a solid bar. 

Strength.—The strength of steel depends upon its chemical 
composition, and upon the mechanical and heat treatment of 
the metal during the process of manufacture. The range of 
strength is very great, the tensile strength alone varying from 
40,000 pounds in a very low steel to 240,000 pounds in steel wire 
employed in cable manufacture. 

The following are considered safe values in engineering 
practice: 


Stress. 

Breaking Unit Stress, 
Pounds 

per Square Inch. 

Allowable Unit Stress, 

Pounds per Square Inch. 

Tension . 

Compression in 

60,000 to 70,000 

8000 (in floor-beams subjected to 
sudden loads) to 16,000 (in 

buildings) 

S v % ' -* 

columns. 

50,000 

10,000 (in bridges) to 16,000 (in 
buildings) 

Shearing. 

Bending. 

48,000 to 56,000 

12,000 

12,500 (in floor-beams' 
of bridges) to 

16,000 (in floor-beams 
of buildings) 

10,000 

20,000 

In c rease 
• 25 percent 
for pins. 

Torsion. 

Bearing. 

60,000 



Unit stresses in wind bracing, about 25 per cent greater. 

Modulus of longitudinal elasticity, 29,000,000 pounds. 

Limit of longitudinal elasticity, one-half of breaking unit stress. 

Average per cent of elongation in 8 inches: ! steel, 22 per cent, 

0 ^ & t low steel, 25 per cent. 

Average per cent of reduction of area: ( J Tle ^^ m steel, 43 per cent 
& r {low steel, 47 per cent. 


DEFECTS. 

The principal defects in steel are due to excess of sulphur 
and phosphorus. The former makes the metal brittle when 
hot, or red-short , and the latter brittle when cold, or cold-short. 
These defects can be avoided only by specifying the maximum 
allowable percentage of each of these elements in the composition 
of the steel. 

















METALS. 


355 


DURABILITY AND PRESERVATION. 

Steel rusts under the action of alternate exposure to dryness 
and moisture, and the action of gases resulting from the com¬ 
bustion of coal; it is protected by covering it with some impervi¬ 
ous coat of oil or paint which is renewed at intervals. The sur¬ 
face to which the paint is applied should be dry and chemically 
clean. At the mill, the metal is cleaned in an acid followed by 
an alkali bath. Old paint is removed from steel structures by a 
sand-blast or by wire brushes. 


SPECIFICATIONS. 


The following specifications are those required by Mr. L. L. 
Buck, Chief Engineer of the Williamsburg Suspension Bridge 
over the East River, New York: 

Structural Steel.—“All steel shall be made in an open-hearth 
furnace lined with silica. 

“The finished steel shall not contain more than the following 
proportions of the elements named: 


Phosphorus 
Sulphur. .. 
Manganese 
Silicon.... 
Copper.... 


.07 of one per cent 


.04 


u 


u 


u 


(( 


U U (( u 

U (C u u 

.10 



.02 


cc 


u 


u 


u 


“ Specimens cut from the finished material shall have the 
following physical properties: 


Description of Material. 

Strength in Pounds 
per Square Inch. 

Elongation 
Per Cent in 

8 Inches. 

Reduction 
Per Cent in 
Area. 

Shapes and universal mill-plates . . 

60,000 to 68,000 

22 

44 

Shear-plates. 

60,000 to 68,000 

20 

44 

Rivet-rods. 

54,000 to 60,000 

25 

50 


“All specimens cut from plates and shapes shall bend cold 
180 0 around once the thickness of the specimen; when at or 
above red heat, 180° flat; and when quenched in water at a 
temperature of 8o° F., 180° around three times the thickness 
of the specimen. Specimens cut from rivet-rods shall bend 






















356 


CIVIL ENGINEERING. 


i8o° flat when cold, when at or above red heat, or when quenched 
from a light yellow heat in water at a temperature of 6o° F. 

“The elastic limit of the steel shall not be less than one-half 
the ultimate strength. 

“All bending tests shall show no signs of fracture on the out¬ 
side of the bent portion. 

“The fracture of all tension tests shall have a cup or angular 
shape and shall have a fine silky texture, of a bluish gray or 
dove color, free from black or brilliant specks. 

“All rolled or forged material shall be entirely free from 
piping, checks, cracks, and other imperfections, and shall have 
smooth-finished surfaces and edges. 

“Rivets cut out of the work, when required by the engineer 
or his representative, shall be tough and show a silky texture 
without a crystalline appearance. 

“Rigid tests will be made to guard against all red-shortness. ,, 

Steel for Castings.—“Steel for castings shall be made in an 
open-hearth furnace lined with silica. 

“The finished steel shall not contain to exceed the following 
limits of the elements named: 

Phosphorus.06 of one per cent 

Sulphur.04 “ “ “ “ 

Manganese.08 “ “ “ “ 

Silicon.35 “ “ “ “ 

'‘‘All steel castings shall be carefully and thoroughly annealed. 

“All castings shall be sound and as free from blow-holes as 
the latest and best practice can produce. 

“Test-pieces taken from coupons on the annealed castings 
shall show an ultimate strength of not less than 60,000 pounds 
per square inch, an elongation of not less than 20 per cent in 
2 inches, and shall bend 90° around three times their thickness 
without rupture. 

“All steel castings must be true to the drawings, with smooth 
surfaces, and all reentrant angles must be neatly filleted. They 
must be planed smooth and true where the drawings require, 
and all holes for bolts must be drilled accurately.” 

Steel for Wire.—“All steel for wire shall be made in an open- 
hearth furnace lined with silica. 






METALS. 


357 


“The finished steel shall not contain to exceed the following 
limits of the elements named: 

Phosphorus 

Sulphur_ 

Manganese 
Silicon. ... 

Copper.... 

Wire. —“The wire for the cables and for the suspenders and 
ties must have an ultimate strength of 200,000 pounds or more 
to the square inch, and must have an elongation under test of 
at least 2J per cent in 5 feet of observed length, and of at least 
5 per cent in 8 inches of observed length. 

“It must be capable of being coiled around a rod of its own 
diameter without cracking.” 

In specifications for less important structures the allowable 
percentages of phosphorus are alone stated; very frequently 
reliance is placed on the mechanical tests alone. 

Wrought Iron. 

Wrought iron is made by puddling in a reverberatory furnace 
certain grades of pig iron , known as furnace pig, which are the 
product of the blast-furnace. From the reverberatory furnace 
the pasty metal is dumped into squeezers where the cinder is ex¬ 
pelled, and then passed through rolls which reduce it to the 
market forms. High grades are obtained by cutting up the 
first or muck bars and repuddling once or twice. Before it was 
replaced by steel in this country the ordinary market forms were 
similar to those of structural steel. 

PROPERTIES. 

Specific Gravity. —The specific gravity of wrought iron is 
from 7.4 to 7.9; in ordinary computations its weight is assumed 
as 480 pounds per cubic foot. 

Expansion and Contraction. —Under the influence of tem¬ 
perature a bar expands and contracts 0.0000067 of its length 
for each degree Fahrenheit; under the action of an applied force 


04 

of 

one 

per 

cent 

°3 

u 

(t 

u 

u 

5 ° 

u 

(( 

u 

a 

,10 

u 

it 

it 

it 

,02 

it 

a 

a 

it 







358 


CIVIL ENGINEERING. 


a bar one inch in cross-section is extended or compressed 
of its length for every ton of applied weight. 

Ductility and Weldability.—Wrought iron is less ductile 
than structural steel, but can be welded more easily. A welded 
joint has also about one-half the strength of an unwelded piece 
of the same area of cross-section. . 

Strength.—In general the strength of wrought iron may be 
assumed to be about 84 per cent of that of soft steel. The fol¬ 
lowing may be considered safe values: 


• 

Breaking Unit Stress, 
Pounds 

per Square Inch. 

Allowable Unit Stress, 
Pounds per Square Inch. 

Tension. 

48,000 

12,000 

Compression. 

48,000 

10,000 

Shearing. 

50,000 

9,000 

Bending. 

48,000 

12,000 (50% more on pins) 

Torsion. 

50,000 

10,000 


Limit of longitudinal elasticity, 26,000 pounds. 

Modulus of longitudinal elasticity, 27,000,000 pounds. 

Percentage of elongation in 8 inches, 15 to 20 per cent. 

Percentage of reduction in area, 12 to 30 per cent. 

Defects.—The defects in wrought iron are also those arising 
from excess of sulphur and phosphorus, red- and cold-short. 

Durability and Preservation.—Wrought iron is durable under 
the same conditions and is protected against rust in the same 
manner as steel. 

Specifications and Tests.—The following specifications for 
wrought iron are given by Fowler in his “ Roof-trusses ”: 

“Wrought iron must be tough, fibrous, and of uniform quality. 
Finished bars must be thoroughly welded during the rolling, 
and be straight, smooth, and free from injurious seams, blisters, 
cracks, or imperfect edges. 

“For tension tests the piece shall have as near one-half square 
inch of sectional area as possible, and a length of at least 8 inches 
with uniform section, for determining the elongation. 

“The elastic limit shall not be less than 26,000 pounds per 
square inch for all classes of iron. 

“Standard test-pieces from iron having a section of 4J square 
inches or less shall show an ultimate strength of not less than 














METALS. 


359 


50,000 pounds per square inch, and an elongation in 8 inches 
of not less than 18 per cent. 

“ Standard test-pieces from bars of more than 4J square 
inches section will be allowed a reduction of 500 pounds for 
each additional square inch of section, provided the ultimate 
strength does not fall below 48,000 pounds, or the elongation in 
8 inches below 15 per cent. 

“All iron for tension members must bend cold through 
90° to a curve whose diameter is not over twice the thickness 
of the piece, without cracking. 

“Not less than one sample out of three shall bend to this 
curve through 180° without cracking. 

“When nicked on one side and bent by a blow from a sledge, 
the fracture must be wholly fibrous.’’ 

JOINTS AND PIECES IN IRON-WORK. 

Joints in iron-work are either lap- and butt-joints with fish¬ 
plates, the pieces being held together by bolts or by rivets, or 
they are eye-bar and pin joints. If several plates are united to 
form a single plate, care is taken to break joints. 

Ties.—Ties are usually made of steel or wrought iron eye- 
bars, the segments being united by pin-joints. If stiffness is 
required in a tie, it is made in the form of a laced or latticed 
column. 

Struts.—Cast-iron struts are in the form of hollow circular 
or square columns. They are employed in structures not sub¬ 
jected to the shocks of suddenly applied heavy live loads. Hori¬ 
zontal beams rest on and are bolted to brackets cast on the col¬ 
umns. Steel and wrought-iron struts are in the form of built- 
up columns made of channels, Z bars, or other convenient struc¬ 
tural shapes. 

Beams.—Steel rolled I beams are manufactured in market 
sizes from a depth of 3 inches and a length of 21 feet to a depth 
of 24 inches and a length of 36 feet. They vary in weight from 
5^ to 100 pounds per linear foot. These I beams may be em¬ 
ployed in pairs, threes, etc., placed side by side to secure greater 
strength. 

When I beams fail to give the requisite strength, plate and 
box girders are constructed as heretofore described. 


3 6 ° 


CIVIL ENGINEERING . 


Other Metals. 

Copper.—After iron and steel, copper is the most important 
metal in engineering practice. Its principal uses are in the 
forms of sheets for roofing, and of wire for electrical purposes. 

Properties.—Metallic copper is both malleable and ductile 
and can therefore be reduced to sheets or wires. The properties 
above mentioned are found in the highest degree in pure copper. 
It is harder and more tenacious than any other metaj excepting 
iron. Its tensile strength is 30,000 pounds, which may be 
doubled by wire-drawing. It is very durable under ordinary 
atmospheric conditions and needs no coating of paint or other 
preservative. Sheathing-copper is formed into sheets 14X48 
inches, whose weight varies from 14 to 34 ounces per square 
foot; these are united by soldering or brazing. 

Tin.—In engineering practice tin is used in the form of tin¬ 
plate, which is sheet iron tinned on both sides; good tin-plate is 
plated with pure tin; the cheaper or “ terne ” plates with a dark 
alloy. The coating of the former is bright, even, and thick, 
and devoid of dark spots or roughness due to imperfections 
in the coating or incomplete coating. 

Zinc.—Zinc, being extremely durable when exposed to the 
action of the weather, is largely used as a coating for sheet iron 
and iron wire. Galvanized iron is made by dipping the sheet 
iron in melted zinc. Tin plate and galvanized iron exposed 
to the action of the atmosphere are usually protected by a coat¬ 
ing of paint. 

Lead.—Lead is used in the form of sheets and pipes. 

Alloys.—Various alloys of these metals are also used to some 
extent in engineering practice. These are described in text-books 
in chemistry. 

For further information consult Thurston’s “Materials of Engineering, 
Parts II and III; Johnson’s “Materials of Construction.” 


CHAPTER XX. 


NATURAL AND ARTIFICIAL STONE. 


Natural Stone. 


Natural stone is a valuable material in engineering practice 
because of its durability and wide distribution; it has a more 
limited field of usefulness than timber because of its great weight, 
and because it is suitable only for resisting crushing and shear¬ 
ing forces. The value of any variety of stone depends upon 
the strength,' durability, appearance, ease of working, and its 
action when exposed to the weather, or to the atmosphere of the 
position in which it is to be used. All of these qualities are more 
or less dependent upon its structure. 

Structure. —According to their predominating constituents, 
stones are classified as silicious, calcareous, and argillaceous; the 
most important constituent of the first being quartz, of the second 
lime, and of the third clay. According to the manner in which 
they are formed they are sedimentary or stratified, igneous or 
unstratified, and metamorphic . 

The ordinary building-stones are given in the following table: 


Class. 

Silicious 

Calcareous 

Argillaceous 


Stratified. Unstratified. Metamorphic. 

Sandstone j 'prap^ 6 } Gneiss 

Common limestone . Marble 

. . Slate 


Stratified and Unstratified. —The stratified stones are found 
in layers varying in thickness from a few inches to many feet. 
As they are easily separated along the planes of stratification, 
they are more easily quarried than unstratified stones. A strati¬ 
fied stone should be laid with its planes of cleavage perpendicular 
to the maximum force to which it is subjected; in this position 
it offers the greatest resistance. It should never be laid with 
its planes of stratification parallel to its exposed surface, since 
in this position it flakes under the action of frost. 


361 





362 


CIVIL ENGINEERING. 


Some of the metamorphic stones, gneiss and slate, have sur¬ 
faces, or planes of cleavage, similar to the planes of stratification. 
They should therefore be treated as stratified stones. 

Unstratified stones may be taken out in blocks, limited only 
by their strength to resist the bending stresses involved in their 
movement; they are therefore largely used in structures where 
massive effects are desired. The metamorphic rocks without 
planes of cleavage, e.g. marbles, are also equafiy strong in all 
directions, and may be treated as unstratified stones. 

Silicious.—The silicious stones are sandstones, granite, syenite,. 
trap, and gneiss. In engineering practice the term granite is em¬ 
ployed as a general term covering granite, syenite, and gneiss. 

Sandstone is formed of grains of sand cemented together by 
silicious, ferruginous, calcareous, or clayey material. The quality 
of the stone depends largely upon the cementing material. The 
granites are of more or less crystalline structure, the quality of 
the stone depending largely upon the size of the crystals and 
the uniformity of the structure. Trap is not properly a building 
material, but is used in engineering practice in road construction. 

Calcareous.—The calcareous stones are either limestones or 
marbles. The limestones are of two general classes, the common 
and the magnesian. The common limestones are of the oolitic 
and the coquina varieties, and are composed of almost pure car¬ 
bonate of lime; the magnesian limestones, or dolomites, contain 
20 to 40 per cent of magnesia. 

The marbles are crystallized limestones containing impurities 
which impart characteristic colors. 

Argillaceous.—The only important argillaceous stone is roofing- 
slate. Bluestone is an argillaceous sandstone employed for curb¬ 
ing and sidewalk slabs. 


PHYSICAL QUALITIES. 

Specific Gravity.—Stones suitable for building construction 
range in specific gravity from 2.2 to 2.8. The unstratified and 
metamorphic stones are heavier than the stratified ones. In 
ordinary computations the weight is assumed as 160 pounds per 
cubic foot. 

Rift and Grain.—The rift of a stone is the direction parallel 
to its plane of stratification or cleavage. Its grain is perpendic- 


NATURAL AND ARTIFICIAL STONE. 363 

ular to this plane. It is usually possible to separate the stone 
in planes in both directions. Paving-blocks are thus made. 

Color.—The chief coloring-matter in stone is iron; therefore 
a very light or nearly white color indicates an absence of iron. If 
the iron is in a form which cannot undergo further oxidation, 
the color is permanent, as in brownstone, in which the iron is in the 
form of red oxide; if, however, it is in the form of a carbonate or 
sulphide , oxidation usually takes place when the stone is exposed 
to the atmosphere and the color of the stone is gradually changed. 
In a building-stone permanency of color is often an important 
consideration. 

Hardness and Toughness.—These qualities enable a stone to 
resist abrasion and concussion. For ornamental work, stones of 
the unstratified or metamorphic varieties, such as fine granite 
and marble, which take a polish, are the most valuable. For 
street pavements, a hard, tough stone of granular structure, such 
as coarse granite or Medina sandstone, is best. Trap, which 
is the hardest and toughest of stones, is very valuable for broken- 
stone roads, but is too smooth for paving-blocks. 

Moisture.—The moisture found in stones in their natural 
state is called quarry-water; this is removed by evaporation when 
the stone is exposed to the air. The quarry-water renders the 
stone softer and is an important factor in working many varieties, 
which become exceedingly hard on exposure. The unstratified 
and metamorphic stones absorb very little water, not over 0.16 
per cent; while the most porous stratified rocks absorb as much * 
as 5.50 per cent. The latter therefore are much less durable 
in a climate where the stone is subjected to a temperature fre¬ 
quently rising above and falling below the freezing-point. On 
this account quarries of porous rock are usually closed during 
freezing weather. 

» 

MECHANICAL PROPERTIES. 

Strength.—The strength of stones to resist a crushing force 
seems to be a function of their specific gravity, since the heavy, 
unstratified, and metamorphic stones are stronger than the lighter, 
stratified ones. From the complex character of their structure 
it is apparent that great differences may be expected in the same 
class. A minimum crushing strength of 5000 pounds per square 


3 6 4 


CIVIL ENGINEERING. 


inch is considered desirable in a building-stone; some granites 
have five times this strength. The stresses per square foot al¬ 
lowed in building regulations are 60 tons in granite, 40 in lime¬ 
stone and marble, and 30 in sandstone. When its resistance 
to crushing is an important feature in the design of a structure, 
as in high reservoir walls and bridge piers, the stone must be 
carefully selected. As stone is lacking in fibrous structure, its 
strength to resist tension and bending is very small compared 
with its strength to resist crushing and shearing. 

DURABILITY AND PRESERVATION. 

Durability.—In engineering practice stones are subjected to 
two classes of disintegrating agencies, chemical and mechanical . 
The principal chemical agents are the acids of the air, notably 
sulphuric and carbonic acids resulting from the combustion of 
gas, coal, etc., in cities and manufacturing districts. Every con¬ 
stituent of stone except quartz is subject to attack by acids. The 
silicious stones having the largest percentages of silica, or the 
quartzites , are therefore the most durable, and the limestones 
the least so. If the cementing substance of a stone is attacked, 
the stone will crumble; if the grains are attacked, a spongy, 
porous stone will result. The durability of a stone, so far as 
chemical agents of destruction are concerned, is chiefly dependent 
upon its mineral constituents. A porous texture will, however, 
expedite the deterioration. Laboratory tests and microscopic 
examinations furnish an indication of the durability of a stone 
under the attack of chemical agents, but the only thoroughly reli¬ 
able test is the exposure of the stone for a long period of time to 
the conditions under which it is to be used. Examination of 
exposed ledges at the quarry shows how well the stone resists 
deterioration under the conditions obtaining there, but is not 
a sure indication of its resisting power under different con¬ 
ditions. 

The mechanical agents to which stones are subjected in engi¬ 
neering practice are the friction and concussion of moving bodies, 
driving winds and rains , and changes of temperature , especially 
in the vicinity of the freezing-point, when they are saturated with 
moisture. The effects of friction and concussion are best seen 


NATURAL AND ARTIFICIAL STONE. 


365 


in the paving-stones of a large city, where granite blocks must be 
replaced every eighteen years, and sandstone in eight. Driving 
winds and rains slowly wear away the surface of stones; the 
action is ordinarily appreciable only in ornamental work where 
sharp edges are exposed to their action. Silicious, unstratified 
stones, which are hard and tough, best resist the above agents. 
The most destructive agent is the range of the temperature in 
temperate climates, which subjects the elements of stone to un¬ 
equal expansion and contraction, opens the joints of masonry 
structures and allows rain to enter, and finally expands this 
water, when the temperature falls below the freezing-point, and 
crumbles the stone. A non-porous stone of uniform structure 
will best resist changes of temperature. If the range of temper¬ 
ature is very great, as in a conflagration in which the heated 
stones are drenched with cold water, the silicious stones of uni¬ 
form texture, as the sandstones, are much more durable than those 
of more complex structure, like the granites. Limestones are 
said to be more durable below the temperature at which the 
stone is calcined. In the Census Report of 1880 the order of 
durability is given as follows: sandstone , granite , marble , ordi¬ 
nary limestone , porous sandstone. 

Preservation.—The preservation of stone may be effected in 
the same manner as wood by coating it with an impervious cover¬ 
ing. The various substances used for this purpose are paint, 
boiled linseed-oil , melted paraffin , and a mixture of alum and soft 
soap dissolved in water. Other processes have been devised for 
filling the pores with an insoluble salt. Ransome’s process con¬ 
sists in saturating the stone with a solution of silicate 0] soda or 
potash and then applying a solution of chloride oj calcium. The 
resulting reaction produces chloride oj soda , which is easily washed 
off, and an insoluble silicate oj lime, which fills the pores of the 
stone. There are several other processes having the same general 
object. 

Quarrying.—Were the stone in every quarry a homogeneous 
monolithic structure, it could be removed from its bed only by 
drilling a row of vertical holes parallel to the face of the quarry 
and a row of horizontal holes parallel to, and at the desired 
distance below, the top. These holes charged with explosive 
and fired simultaneously would loosen the stone from its bed 


366 


CIVIL ENGINEERING. 


when it could be broken into smaller pieces in a similar manner. 
This process is abridged in many quarries due to the planes of 
stratification and cleavage, and also due to the joints which divide 
the mass of stone into volumes of different dimensions. Strati¬ 
fied and foliated stones do not, as a rule, require the process of 
undercutting, or gadding , as the stones are easily separated along 
the planes of stratification and cleavage. They are also divided 
by joints perpendicular to these planes. Unstratified rocks are 
also usually traversed by two or more sets of intersecting joints 
which divide them into more or less regular rectangular or 
rhomboidal prisms. The planes of stratification and cleavage 
and the joints limit the size of the product which may be 
obtained from any quarry, but also decrease the cost of the 
quarried stone. 

The drills used in quarrying are hand-, churn-, or steam- 
drills. Hand-drills are short blunt chisels which are forced into 
the stone by blows from a hand-hammer or sledge. Churn-drills 
are longer drills which are forced into the stone by dropping them 
a short distance. They are suitable for holes from 3 to 5 feet deep. 
In steam or compressed-air drills the man-power of churn-drills is 
replaced by one of those agents. The blows are delivered with 
great rapidity. In quarries of sandstone, limestone, and marble 
a channeling-machine is often employed instead of the steam or 
compressed-air drills. This machine operates in a similar man¬ 
ner, but cuts a continuous groove instead of a series of holes. A 
hollow revolving drill or diamond drill is employed when large 
holes are required; this drill removes a core and thus furnishes 
specimens for the study of the character of the stone. Explosives 
employed in quarrying dimension-stone are slow-acting and used 
in small quantities. The direction of the line of fracture is in¬ 
fluenced by the form of the hole around the charge. Large, 
irregular fragments are obtained by the use of slow-acting ex¬ 
plosives without any attempt to control the direction of the line 
of fracture. Small irregular fragments are obtained by the use 
of high explosives. 

Dressing.—The quarried stone is reduced to proper dimen¬ 
sions for use by employing hand-drills , plugs and feathers; it is 
then dressed with the ordinary mason’s tools. Rock-face finish 
presents a rough, undressed surface, which is frequently sur- 


NATURAL AND ARTIFICIAL STONE. 367 

rounded by a margin of drove-work; the latter is made with a 
wide chisel or drove. Pointed-jace finish is rock-face finish 
trimmed down with a sharp-pointed tool called a point. Pean - 
hammered or ax-hammered finish is a comparatively smooth 
surface made by hammering the surface with a pean-hammer; 
this is a heavy axe with a blunt cutting-edge. Patent-hammered 
finish is similar to the pean-hammered finish and is made by 
hammering the surface of the stone with a hammer whose head 
is made of several plates of steel so fastened together as to form 
a grooved face. The finish is known as 4-cut, 6-cut, 8-cut, 
10-cut, and 12-cut, depending upon the number of plates per 
lineal inch in the hammer. This is the ordinary finish of smooth- 
surface walls. Bush-hammered finish is similar to the last, the 
hammer-head being a single piece with a grooved face. Tooth- 
chiseled finish is made with a saw-toothed chisel; the softer 
stones are sometimes finished in this manner. Sawed jace is, as 
its name implies, produced by sawing the stone. Fine sand 
and pumice finish are made by rubbing the surface with fine 
sand or pumice. A polished surface is then given by rubbing 
with polishing-putty , or a moist woolen cloth and oxalic acid. 

Specifications.—Specifications for building-stones usually re¬ 
quire the bidder to submit samples for appearance and samples 
for tests. The former arc six-inch or one-foot cubes whose 
faces are dressed in the different styles required in the structure. 
Two-inch cubes or test-pieces are made with smooth faces. 

Tests.—The tests to which specimens of building-stones may 
be subjected are chemical and mechanical. The chemical or 
laboratory tests are made for the purpose of determining the 
composition of the stone and the probable effect of its various 
elements when exposed to the conditions to which it will be 
subjected in the structure. The mechanical tests are for the 
purpose of determining the amount of absorption, its action 
when exposed to frost in a saturated condition, its action when 
exposed in a conflagration, and its crushing strength. The 
absorption test consists in drying the specimen in an oven and 
then immersing for a long time in water; after it is removed from 
the water it is exposed to the air until the surface-water is evapo¬ 
rated. The difference in weight before and after immersion will 
give the weight of the water absorbed. 


368 


CIVIL ENGINEERING. 


The jreezing test consists in exposing a specimen saturated 
with water to the alternative action of a temperature above and 
below the freezing-point. The test may be made in summer 
by exposing the specimen to the temperature of the air each day, 
and the temperature of a refrigerating apparatus each night for 
a long period of time. The total loss of weight will measure the 
action of the frost. The fire test consists in exposing it to high 
heat and dashing it with cold water. The crushing strength is 
determined by crushing the specimen in a testing-machine. 

Brick. 

Brick is an artificial stone which is cheaper than natural 
stone, almost equally strong, and is less affected by the ordinary 
chemical and mechanical agents of destruction. Its field of 
usefulness is, however, limited by the small size of the individual 
stones, which ordinarily have a volume of only one-twentieth of a 
cubic foot. 


MANUFACTURE. 

Clay.—The clay of which common bricks are made consists 
principally of silicate oj alumina. Sand , lime, iron , and mag¬ 
nesia are usually present in varying proportions. Sand if not in 
excess is beneficial, as it preserves the form of the brick when ex¬ 
posed to the high temperature of burning; in excess it destroys 
the cohesion and renders the brick brittle and weak. Silicate 
of lime renders the brick fusible and liable to deformation in 
burning. Carbonate of lime in excess is calcined in the process 
of burning, slakes in contact with moisture, and causes the brick 
to crumble. Iron, which in burning forms the red oxide, gives 
the brick strength, hardness, and a red color. By selecting the 
clay and introducing coloring material bricks of many colors 
and shades of color are produced. 

Mixing.—The clay is converted into a plastic mass for mold¬ 
ing by mixing with water and kneading In small plants the 
mixing is effected by shoveling the clay into a circular trough, 
where it is moistened with water, and broken up by a heavy 
wheel which moves around the trough, or by mixing the clay 
and water in a pug-mill. The latter is a vertical cylinder or 
box in whose axis is a vertical revolving shaft. To this shaft 


NATURAL AND ARTIFICIAL STONE. 


369 


are attached horizontal arms set in such a manner as to knead 
the clay and force it downward and out through an orifice in the 
bottom. 

Molding. — The paste is molded by throwing it into plank 
molds, open at the top and bottom, which are placed on a mold¬ 
ing-board. The mold is carefully filled and struck off with a 
straight-edge. To prevent the paste from adhering to the sides 
of the molds they are kept wet and sprinkled with sand. In 
large plants the operations of mixing and molding are done by a 
continuous process in a brick-making machine. In one form 
the plastic clay passes from the mixer into a hopper from which 
it is forced through a rectangular aperture whose cross-section 
is that of a brick; here it is cut by a wire to the proper length 
and is carried off on an endless chain to the drying-floor. In 
another form it drops from the hopper into molds attached to the 
circumference of a large wheel; from these molds the bricks are 
also deposited on an endless chain. 

Classification.—Common brick are classified according to the 
way in which they are molded, into hand-made , machine-made , 
and pressed brick. Machine-made brick are usually more regu¬ 
lar than hand-made brick, but some machines injure the struc¬ 
ture of the brick itself. Pressed brick are very regular, compact 
bricks, produced by compressing a partially dried brick in a 
suitable mold under heavy pressure. In the old style of kiln, 
brick are classified as overburned or arch brick , cherry or hard 
brick, and salmon , pale , or sojt brick. The first are hard and 
brittle, and do not form a good bond with mortar; the last are 
too soft for general use and are used only in interior walls sup¬ 
porting light weights. 

Requisites of a Good Brick.— x\ good brick should have 
plane faces, parallel sides, and sharp edges and angles; it should 
be of fine, compact, uniform texture, should be quite hard, and 
give a clear, ringing sound when struck. It should not absorb 
more than one-tenth of its weight of water. Its specific gravity 
should be at least two. Its crushing strength should be at least 
7000 pounds per square inch, and its modulus of flexure at least 
1000 pounds per square inch. 

Tests.—In purchasing bricks it is best to specify that the 
bidder shall submit samples representing the worst of the brick 


37o 


CIVIL ENGINEERING. 


to be furnished. The qualities above enumerated are determined 
as follows: The shape and structure are determined by 
careful inspection of the entire brick, and its cross-section when 
broken. The amount of absorption is determined as in the 
test for stone. The conditions of practice are best represented 
by testing whole bricks, but, as some kinds have a more or less 
impervious skin, it is well to make a similar test on a broken 
brick to determine the condition of the interior. The specific 
gravity, the crushing and bending strengths are determined as in 
tests of other materials. 

Size and Weight.—There is no legal standard of size in the 
United States. In the Eastern States the usual size is 8JX4X2J, 
but in the West the dimensions are somewhat smaller. On 
account of the shrinkage in burning, the size is not exactly uni¬ 
form. Where uniformity of size is essential it is usual to specify 
that the brick shall be culled , i.e., assorted in sizes. Pressed 
brick weigh about 150 pounds per cubic foot, hard brick about 
125 pounds, and inferior soft brick about 100 pounds per cubic 
foot. Common brick average about 4J pounds each. 

SPECIAL FORMS. 

Paving-bricks.—Paving-bricks are used for street pavements, 
stable floors, etc.; they are ordinarily made of clay in the form 
of shale, which has a higher percentage of flux than common 
brick-clay, and is burned at a higher temperature than common 
brick. The result is a compact semi-vitrified brick or paving- 
block. A good paving-brick should be hard, tough, and absorb 
little water even when its surface is worn off. The hardness 
may be tested by grinding the brick on a stone, and the tough¬ 
ness by placing the brick in a rattler used for cleaning castings, 
or by dropping it repeatedly on a hard floor. 

Fire-bricks.—Fire-bricks are used where high temperature is 
to be resisted. They are made of nearly pure clay, or of a mix¬ 
ture of nearly pure clay and clean sand or crushed quartz. The 
presence of oxide of iron is injurious. Fire-bricks should con¬ 
tain less than 6 per cent of oxide of iron, and less than an aggre¬ 
gate of 3 per cent of combined lime, soda, potash, and magnesia. 
Good fire-brick should be uniform in size, regular in shape, 
easily cut, strong and infusible. 


NATURAL AND ARTIFICIAL STONE. 


371 


Terra-cotta.—The term terra-cotta is usually applied to brick 
which has been molded into ornamental forms. It is made of 
different varieties of selected clays mixed together with ingredients 
which make the mixture slightly fusible. The forms are given by 
plaster molds. 

Tiles.—The term tile is applied to a variety of forms of brick 
employed in building construction. The dense tiles employed 
in fire-proof floor construction are made of fire-clay mixed with 
other clays, and are shaped in strong molds under heavy pres¬ 
sure. Porous tiles are made in a similar manner; with the 
clay is mixed some sawdust or straw which is reduced to ash 
in the burning. Roofing-tiles are dense tiles often made with 
a glazed surface. 

Glazed and Enameled Bricks.—These are special varieties 
of brick having a glazed or enameled surface. The surface may 
be given any desired color. 

Vitrified Sewer-pipe and Blocks.—Sewer-pipe is made of 
selected clay and has a glazed coating. The pipes are molded 
by machinery, dried, placed in a kiln, and gradually exposed to 
a high heat. At the proper temperature coarse salt is thrown 
on the fire; the salt vaporizing combines with the silica of the 
clay and forms a soda-salt, or glass. Sewer-blocks are similar 
in composition to sewer-pipes; they are used to replace the 
bottom courses of a brick sewer. 

Lime, Cement, Mortar, and Concrete. 

Lime and cement are the products of the burning or calcina¬ 
tion of certain classes of calcareous stones. The receptacle in 
which the operation takes place is called a kiln. The character 
and properties of the product of calcination depend upon the 
chemical composition of the limestone and the temperature at 
which the stone is burned. These products are common lime } 
hydraulic lime , and cement. Any one of them if reduced to pow¬ 
der and mixed with water will form a paste which is the cement¬ 
ing material of a large class of artificial stdnes. If the paste is 
mixed with sand only, the resulting mixture is called mortar . 
Mortar is itself employed either as a paste to unite individual 
stones in masonry, or it is employed as an artificial stone. Con - 


372 


CIVIL ENGINEERING. 


Crete is an artificial stone usually made by mixing cement mortar 
in the plastic state with broken stone, gravel, etc. 

Liquid asphalt and coal-tar are also employed in the place 
of lime or cement Daste in making mortar and concrete. 


LIME. 

Common Lime.— Rich , jat, or pure lime is produced from 
limestone which is practically pure calcium carbonate; marble, 
white chalk, oolitic and coquina limestone, in which the im¬ 
purities do not exceed 2 or 3 per cent, may be employed for this 
purpose. The calcination at moderate heat drives off the car¬ 
bonic acid and water in the stone and leaves quick or caustic lime. 
This product is white, amorphous, highly caustic, and has great 
avidity for water. If lumps of quicklime are sprinkled with 
water, they will at once swell, burst into small fragments, and 
finally crumble into a powder known as slaked lime or calcium 
hydrate. The process of slaking is accompanied by the evolu¬ 
tion of heat and steam, and by an increase in volume, varying 
between two and three and a half times the original volume. 
The same effects are produced without the evolution of heat and 
steam if the quicklime is exposed to the atmosphere; it absorbs 
moisture slowly and slakes gradually; this is known as air-slaking. 

If a pat made of slaked lime and water is exposed to dry air, 
it will crack due to contraction, and the outside will harden, due 
to its absorption of carbonic acid and conversion into calcium 
carbonate. If the mass is large, the interior will become friable 
in dry air, and remain pasty in damp air. If a similar pat is 
placed under water, it will disintegrate, and if the quantity of 
water is sufficient to dissolve it, will wholly disappear. 

Meager or poor lime is produced by the calcination of lime¬ 
stone composed of calcium carbonate mixed with sand. The 
process of slaking is retarded and rendered less complete by 
the sand; if the lime is very poor, it cannot be reduced to a powder 
by slaking, but must be thus reduced by grinding. Poor lime 
should not be used in engineering work. • . 

Manufacture.—The kiln in which the limestone is calcined is 
a large cylindrical, conical, or egg-shaped chimney which is 
ordinarily lined with fire-brick. The process may be either 
intermittent or continuous. If intermittent, the bottom of the 


NATURAL AND ARTIFICIAL STONE . 


373 


kiln forms the fuel-chamber, which is separated from the lime¬ 
stone chamber above it by a perforated arch of limestone or fire¬ 
brick; this arch supports the limestone and allows the heat 
to pass through and reach the stone in the upper chamber. The 
stone is introduced at the top of the kiln, and in some kilns also 
through doors in the walls. When the charge has been fully 
burned, it is removed through a door just above the arch, and 
the kiln is allowed to cool sufficiently to allow the workman to 
put in a new charge. 

If the process is continuous, one of the two following forms 
of kilns is employed. In the first form the fuel-chamber is con¬ 
structed on the circumference of the limestone chamber near its 
base; the stone is put in at the top and drawn out at the bottom, 
being calcined in that part of the kiln where the heat from the 
fuel-chambers passes into the chamber containing the stone. In 
the second form, the coal and limestone are placed in the kiln in 
alternate layers, the whole charge being supported by a grate 
which is placed across the kiln near its base. The coal in the layer 
at the grate is ignited, and the layer of limestone above it is cal¬ 
cined and removed through the grate The fire is communicated 
from the first to the second layer and thus the process becomes 
continuous. As each layer of coal is consumed it is replaced by 
a new layer introduced at the top, and as each layer of limestone 
is removed it is replaced by a new layer of stone, also introduced 
at the top. The kiln in which the fuel is kept separate from the 
stone is more easily managed and gives more uniform results than 
the one in which the fuel and stone are placed in layers; it re¬ 
quires, however, a greater expenditure of fuel. 

Preservation.—To protect quicklime, slaked lime, and lime 
paste from prematurely absorbing moisture and carbonic acid 
from the atmosphere, they should either be used at once, or they 
should be kept in closed vessels. In building operations it is 
customary to use freshly burned lime which, after being slaked, 
is made into a paste and covered with sand until used. 

Tests.—Good lime should be clean, slake readily, and dis¬ 
solve in water. 

Hydraulic Lime.—Hydraulic lime is produced from certain 
varieties of silicious and argillaceous limestone, notably those 
of Teil and Seilly, France. It is calcined in the same manner 


374 


CIVIL ENGINEERING. 


as common lime and is at once reduced to a powder by slaking • 
it slakes less readily than poor lime, and its increase in volume does 
not exceed one-third its original volume. After slaking it is 
packed in sacks or barrels and protected from the air until used. 

A pat made of the slaked hydraulic lime and water will harden 
not only in the air, but also joinder water. This hardening, called 
setting, differs from that of common lime in taking place through¬ 
out the pat and not simply on the surface. The setting is assumed 
to be due to the crystallization of the hydro-silicates and hydro- 
aluminates formed when the slaked lime is mixed with water; 
the crystals bind the material firmly together. 

Hydraulic lime is not manufactured in this country, as the 
limestones are more suitable for the manufacture of cement. 

Artificial hydraulic lime may be made by mixing soft chalk 
or slaked lime with the proper proportion of clay, molding the 
mixture into bricks, and burning the bricks in a kiln. It is re¬ 
duced to a powder by grinding. 

CEMENT. 

Cement is the product of the calcination of certain natural 
or artificial combinations of calcium carbonate with alumina, 
carbonate of magnesia, and silica; iron and other elements in 
small ratios are also usually present. The product of calcination 
differs from lime in that it cannot be slaked, but must be reduced 
to a powder by grinding. A pat made of cement powder and 
water will set under water as well as in the air, and will increase 
in strength with age. 

Cements are divided into three classes: natural cement, port- 
land cement, and puzzolans. 

Natural cement is that produced bv calcining natural-cement 
rock at a heat below that of incipient fusion. The terms quick- 
setting, light, Roman, American, and Rosendale are all applied 
to this variety. It is usually of a brown color and has a specific 
gravity of 2.7 to 3.0. 

Portland cement is that produced by calcining up to the point 
of incipient fusion intimate mixtures, in exact proportions, either 
natural or artificial, of carbonate of lime and clay. Its name 
is derived from its resemblance in color to portland stone; the 


NATURAL AND ARTIFICIAL STONE . 


375 


terms heavy and slow-setting are also applied to this cement. 
It is of a gray color and has a specific gravity of 3.10 to 

3 - 2 5 - 

Puzzolan cement is an artificial cement made by mixing in 
proper proportions slaked lime and granulated blast-furnace 
slag. The mixture is not calcined. The name is derived from 
puzzolana, a naturally burned earth of volcanic origin, found 
at Pozzuoli near Naples, Italy. This earth and slaked lime 
will give a mixture having hydraulic properties. Puzzolan cement 
is of a lilac tint and has a specific gravity of 2.7 to 2.8. 

Silica or sand cement is made by mixing portland cement 
with about an equal amount of sand and grinding the mixture 
to a fine powder; it is almost as strong as the neat cement. 

Manufacture.—Natural cement, being made from the stone 
itself, is burned in kilns similar to those described for the calcina¬ 
tion of lime; the process, however, requires a higher temperature. 
After being burned the stone is crushed and finally ground into 
fine powder in suitable machinery. 

Portland cement usually results from the burning of a mechani¬ 
cal mixture, hence the first operation is to form this mixture. 
The processes of manufacture are known as the wet and the dry 
processes. 

In the wet process only soft materials, such as chalk and 
clay, are employed. These are dumped in the proper propor¬ 
tions into a wash-mill with a large amount of water, and the ma¬ 
terials are intimately mixed and reduced to fine powder by means 
of knives or tines attached to horizontal arms fastened to a ver¬ 
tical revolving shaft. The liquid, or slurry , is allowed to flow 
from the wash-mill through sieves into the settling-basins, or backs, 
where the water is drawn and evaporated from the surface. 
When the slurry reaches the consistency of soft butter it is re¬ 
moved with shovels and placed on the drying-floor, where it is 
artificially dried, usually with the waste heat from the kilns. 
When the slurry is sufficiently dry it is burned in a kiln, usually 
of the continuous type, in which the fuel and stone are in con¬ 
secutive layers. The product of the kiln is a very hard and 
heavy dark green honeycombed clinker, which is reduced to 
small fragments in a crusher similar to those used in crushing 
stone. It is finally ground to a powder by millstones. This 


37^ 


CIVIL ENGINEERING. 


process is employed in England, where the materials are chalk, 
and clay from the banks of the Thames and Medway. 

The wet process is also modified into the semi-wet. In this 
the settling-basins are omitted and less water is used in the wash- 
mill. The wet slurry passes from the wash-mill to millstones, 
where it is ground and the mixing of the materials thus com¬ 
pleted. This is also employed in England. 

The dry process is employed where the limestone and clay 
are so hard that they cannot be mixed in a wash-mill. The 
materials are therefore reduced to a powder separately by 
crushing and grinding, and are then mixed in the proper pro¬ 
portions in a pug-mill. From the pug-mill they pass to a brick- 
machine, where they are made into bricks for the kiln. The 
process of burning the bricks and reducing the clinker to powder 
may be conducted as in the wet process. The dry process is em¬ 
ployed in Germany and in the United States. 

The kiln generally employed in this country is the rotary 
kiln. This is a wrought-iron cylinder 60 feet long and 6J feet 
in diameter, which is lined with refractory material. The cylin¬ 
der is slightly inclined to the horizontal and rotated by proper 
machinery. The powdered mixture is introduced either wet 
or dry in a continuous stream at the high end of the cylinder, 
and is reduced to clinker by the flame of powdered coal or gas 
which enters under pressure at the low end. The clinker finally 
drops out at the low end, and the process is thus made continu¬ 
ous. 

The manufacture of portland differs from that of natural 
cement, in that the proportion of ingredients is much more 
exact, the temperature is higher, and the product of calcination 
is harder and less easily reduced to powder. 

In the manufacture of puzzolan or slag cement, the slag from 
a blast-furnace is suddenly cooled by means of a jet of cold water 
and is granulated in a crusher. The lime is prepared in the 
usual manner, and is slaked before it is mixed with the slag. 
After mixture the cement is reduced to a powder by grinding. 

Preservation.—Cement must be protected from contact with 
moisture which will produce premature setting; it is usually 
packed in barrels or sacks lined with stout paper. These must 
be stored under cover on a raised floor. 


NATURAL AND ARTIFICIAL STONE. 


377 


Tests.—The process of manufacturing cement consists of 
three distinct operations: the selection or mixing of the ingredi¬ 
ents, the burning of the stone or mixture, and the reduction of 
the product to powder. Upon the manner in which each of 
these operations is performed depends the quality of the cement. 
These qualities can be determined only by making suitable tests. 
The tests recommended by the Engineer Department of the 
Army for portland and puzzolan cements are tests for sound¬ 
ness, fineness , tensile strength , specific gravity , and for activity; 
in testing natural cement, the tests for soundness and for specific 
gravity are omitted. 

Soundness.—Unsoundness in cement is usually due to an 
excess of free lime resulting from improper proportions or from 
underburning. This free lime expands on contact with water 
and causes disintegration. In this test neat-cement paste is 
made into pats, which are small cones or spherical segments 
3 inches in diameter and a half-inch thick at the center. Faija’s 
test consists in suspending a freshly made pat for six hours over 
water kept at a temperature of 115 0 to 120° F.; at the end of 
this period the pat is lowered into the water, where it is allowed 
to remain 18 hours. The pat when taken out of the water should 
show no signs of disintegration. In the Engineer Department 
tests two pats are made on glass plates, and are exposed to the air 
for 24 hours under a damp cloth. One is then placed in water 
which is gradually raised to the boiling-point, and kept there for 
6 hours, and the other is immersed in fresh water for 28 days. 
The pats should remain firmly attached to the plates, and should 
show no signs of cracking or expanding or otherwise disintegrating. 

Fineness.—All other conditions being the same, the adhesive 
power or the cementitious value of cement, that is, its power to 
bind together particles of inert sand, depends upon its fineness. 
A larger percentage of sand may therefore be employed in making 
mortar with a fine cement than with a coarse one. The En¬ 
gineer Department requirements are.* 

Natural cement 8°/c j s i m ]i p ass through a sieve made of wire 0.005 inch in 

Portlan cement 9 2 To f diameter, having 10,000 openings per square inch. 
Puzzolan cement 97% ) ’ 6 

Fineness may result from the clinker being underburned and 
therefore more easily reduced to a fine powder; this defect will 
be developed by the other tests. 


37 8 


CIVIL ENGINEERING. 



Tensile Strength.—As in the case of metals, the test for 
tensile strength is easily made and furnishes reliable data for 

determining the value of any given cement, and 
for comparing the qualities of different varieties. 
For the purpose of making this test the cement, 
either neat or mixed with sand, is made into a 
paste by the addition of water, and is then 
molded into briquettes. The briquette when 
taken from the mold is of the form shown in 
Fig. 117 and is 1 inch in thickness. Its least 
dimension of cross-section is 1 square inch. To 
secure comparable results it is necessary to employ constant pro¬ 
portions, by weight, of cement, sand, and water in making the 
paste for each briquette, and'to employ the same amount of pres¬ 
sure in forcing them into the mold. 

The percentage of water by weight in neat tests is given by 
the Engineer Department specifications as follows: For natural 
cement 30 per cent, for portland cement 20 per cent, for puzzo- 
lan cement 18 per cent. 

In mortar briquettes the proportions are 1 cement to 1 sand 
in testing natural cement, and 1 cement to 3 sand in testing 
portland and puzzolan cements. In making the mortar the 
water is added after the cement and sand are thoroughly mixed 
in a dry state. The percentage of water, by weight, to the mixed 
cement and sand is: for natural cement 16 per cent, for portland 
12 per cent, and for puzzolan 10 per cent. 

After the briquette is made it is covered with a damp cloth 
and left for 24 hours; it is then removed from the mold and 
immersed in fresh water until tested. The temperature of the 
water should be between 50° and 70° F. The tensile test is made 
in a machine which grips the briquette as shown in Fig. 117, and 
ruptures it at the minimum cross-section by a gradually increas¬ 
ing pull. Several varieties of these machines are in the market. 

The Engineer Department requirements for briquettes which 
are kept 24 hours under a damp cloth and then immersed in 
fresh water are: 

For Natural Cement.—A neat-cement briquette must at the 
end of 7 days have a tensile strength of 90 pounds per square 
inch; at the end of 28 days, a tensile strength of 200 pounds. 






NATURAL AND ARTIFICIAL STONE. 


379 


A briquette of i cement and i sand must at the end of 7 
days have a tensile strength of 60 pounds, and at the end of 28 
days a tensile strength of 150 pounds. 

For Portland Cement.—A briquette of neat cement must 
at the end of 7 days have a tensile strength of 450 pounds per 
square inch, and at the end of 28 days have a tensile strength 
of 540 pounds. 

A briquette of 1 cement and 3 sand must at the end of 7 days 
have a tensile strength of 140 pounds, and at the end of 28 days 
a tensile strength of 220 pounds. 

For a special quick-setting portland the required strength 
is reduced 12 per cent for the neat cement, and 15 per cent for 
the cement and sand briquettes. 

For Puzzolan Cement.—A neat-cement briquette must at the 
end of 7 days have a tensile strength of 350 pounds, and at the 
end of 28 days a tensile strength of 500 pounds. 

A briquette of 1 cement and 3 sand must at the end of 7 
days have a tensile strength of 140 pounds, and at the end of 
28 days a tensile strength of 220 pounds. 

The stress is applied at the rate of 400 pounds per minute; 
the highest result from a set of briquettes made at the same 
time is the governing test. A dozen briquettes of the same 
kind are usually made for testing purposes; the paste is mixed 
for not more than four at a time, lest the cement set before the 
last briquette is molded. 

Specific Gravity.—This test is recommended for the purpose, 
of determining whether a portland cement has been adulterated, 
the materials employed in this process having a less specific 
gravity than properly burned cement. It is also a test for im¬ 
proper burning. 

Activity.—To secure uniform results, it is usual to reject all 
cements which set either too rapidly or too slowly. For this 
purpose two limits are established, known respectively as the 
initial and the final set. The time of initial set is when the neat- 
cement paste will just bear without indentation a wire - r V of 
an inch in diameter, supporting a weight of } pound. The 
time of final, permanent, or hard set is when the paste will just 
bear without indentation a wire -fa of an inch in diameter, 
supporting a weight of 1 pound. 


3 So 


CIVIL ENGINEERING. 


The Engineer Department requirements are: 


Natural cement 
Portland ‘ ‘ 
Puzzolan ‘ ‘ 


Not less than 20 min. 


Initial Set. 


Final Set. 

Not more than 4 hrs. 



45 


“ “ “ 10 “ 
“ “ “ 10 “ 


Quick-setting portland. . . Betw. 20 and 30 “ “ “ f to 2 \ hrs. 

The tests are applied to two pats similar to those employed 
in the test for soundness. 

Short-time Tests.—The boiling test for soundness and the 
tests for time of setting should be made when either time or 
appliances are lacking for the more complete ones. Masons 
test the time of final setting by the pressure of the thumb-nail; 
if no indentation can be made, it is assumed that the cement has 
attained its final set. When only short-time tests can be made, 
the Engineer Department limits the choice to cements which 
have been satisfactorily employed in the locality of the proposed 
structure for at least three years. 


Sand, Gravel, and Broken Stone. 


Sand.—Sand is the granular product arising from the dis¬ 
integration of rocks. Bank or pit sand is that from inland exca¬ 
vations; river and sea sand are those excavated along the shores 
or dredged from the bottom of bodies of fresh and salt water. 

Specifications for engineering work usually require the sand 
to be “clean and sharp” 

By a clean sand is meant one which, if shaken with water 
in a bottle and then allowed to settle, will leave no scum on the 
surface of the water and no layer of fine mud on the surface of 
the sand. River and sea sands, if not affected by the sewage 
of cities, are usually clean. Sea sand, however, contains salts 
which attract moisture and produce an efflorescence on the sur¬ 
face of the masonry; if this is objectionable, the sand must be 
washed in fresh water. A small percentage of non-decaying 
impurity, about 5 per cent, is usually allowed in sand; the per¬ 
centage may be readily determined by shaking the sand with 
water in a graduated glass or bottle, and comparing the depth of 
the surface layer of mud with that of the clean sand. Dirty 
sand may be made clean by washing it in a trough of running 
water. 





NATURAL AND ARTIFICIAL STONE. 381 

By a sharp sand is meant one composed of rough, angular 
grains. Silicious bank sand is usually sharp; river and sea 
sands have rounded grains. The sharpness may be determined 
by rubbing it in the hand or by magnifying it with a lens. Sand 
which is otherwise acceptable is not ordinarily rejected for want 
of sharpness. 

Sand may be classed as coarse if it is retained on a sieve hav¬ 
ing 20 wires to the inch; it may be classed as fine if it passes 
through a sieve having 30 wires to the inch. Coarse sand is 
preferable to fine or mixed, if there is sufficient paste to fill the 
voids. The strength of the mortar depends upon the amount 
of paste. If less paste is used, a mixed sand is better than one 
composed of grains of uniform size, since the small grains will 
assist in filling the interstices or voids between the large ones, 
and thus decrease the amount of paste necessary to make a 
homogeneous mass. 

The dust formed in crushing stone which passes through a 
No. 4 sieve may be employed in the place of sand; it gives a 
stronger mortar than sand itself. 

Gravel.—Gravel is an aggregation of small rounded stones 
varying in diameter from \ to ij inches. It is employed alone, 
and also mixed with broken stone, in making concrete. 

Broken Stone.—Broken stone is the product produced by 
breaking quarry stone in a stone-crusher, and separating it from 
the dust and large fragments, by passing it through an inclined 
cylindrical revolving screen. Circular holes of different sizes in 
this screen first remove the dust and then allow the broken stone 
to fall into bins placed beneath. It is usually specified that the 
stone shall pass through a two-inch ring. The best stone for this 
purpose is tough and breaks into irregular cubes with rough 
surfaces. For concrete the cubes should be of different sizes, so 
that the small pieces may fill the voids of the large ones. 

Lime and Cement Mortar. 

Mortar may be employed not only to bind the parts of a 
masonry structure together, but also as a surface coat of floors, 
sidewalks, etc. The binding material of the mortar is the paste 
made of lime or cement, and water; the sand is simply an inert 
substance, which helps to fill the voids and joints without reduc- 


382 


CIVIL ENGINEERING. 


ing the strength of the binder below the desirable limits, but 
greatly diminishes the cost of the masonry 7 . In common lime 
mortar it also prevents undue shrinkage and increases the com¬ 
pressive resistance of the paste. 

Common Lime Mortar.—The process of making lime mortar 
consists in reducing quicklime to a paste by slaking and then 
mixing this paste with sand. The quicklime, broken into lumps 
of convenient size, is placed to the depth of 6 or 8 inches in a 
water-tight box. With buckets or a hose it is then drenched 
with just sufficient water to insure its thorough slaking or reduc¬ 
tion to powder. During the slaking it may be covered with can¬ 
vas or a layer of sand to retain the heat and increase its activity, 
or it may be stirred for the same purpose. When the lime is 
completely slaked it is covered with the proper volume of sand, 
evenly spread, and the two ingredients are mixed with a shovel 
or hoe until every grain of sand is covered by lime paste. 

If a box is not available, the sand is placed on a platform and 
the lime is put in a basin made in the sand. 

The slaking is the only operation requiring special attention. 
If too much water is employed, the lime paste is too thin; if too 
little water is used, the paste will be full of pieces of unslaked 
lime. The latter defect may be partially remedied by the addi¬ 
tion of more water, but the paste will not be as smooth as it would 
have been had the proper amount of water been used at the 
. beginning of the process. The proper amount of water is de¬ 
termined by experiment. 

Mortar may be mixed in a pug-mill or in a concrete-mixer. 
The volume of sand employed depends upon the richness of the 
lime and varies between 2J and 4 times the volume of lime paste. 

Lime mortar is commonly used as soon as made; as it hardens 
slowly, however, this is not absolutely necessary. 

The use of lime mortar is confined to dry places where it is 
exposed to the air. It is employed in the construction of thin 
brick walls above ground and in the foundation coats of plaster; 
it loses its binding properties when exposed to dampness, as in 
basement walls, and when excluded from contact with air, as in 
thick walls. 

Cement is often added to lime mortar to give it strength. 
The proportions may then be 1 measure lime, 1 cement, 6 sand. 


NATURAL AND ARTIFICIAL STONE. 


3 8 3 


Cement Mortar.—As the cement is in the form of powder, it 
is only necessary to mix it thoroughly with the proper proportions 
of sand and water. There being at least twice as much sand as 
cement, in mixing, one half the sand is spread in an even layer 
in a box or on a platform; the cement is spread over this, and 
the remainder of the sand over the cement. The two ingredi¬ 
ents are then thoroughly mixed by turning the mass over at least 
three or four times with a shovel. A bowl-shaped depression 
being made in the center of the mass into which the proper amount 
of water is poured, the process of mixing is then repeated until 
every grain of sand is coated with cement paste. The strength 
of the mortar depends largely upon thorough mixing. Cement 
mortar may also be mixed in a pug-mill or a concrete-mixer. The 
proportions commonly employed are i measure of natural cement 
to 2 of sand, and i measure of portland or puzzolan cement to 
3 of sand. Both stronger and weaker mixtures are also used. 
A perfect mortar is one in which the cement exactly fills the voids 
in the sand. The volume of voids varies from 30 to 50 per cent 
of the volume of the sand. 

Cement mortar must be used immediately after mixing so as 
to avoid the setting of the cement before the mortar is in place; 
it is sometimes specified that the mortar must be in place one hour 
after it is made. 

Uses.—Portland-cement mortar is employed in structures in 
which great strength is required, as in masonry dams and masonry 
arched bridges; where the surface is exposed to mechanical 
wear, attrition, or blows, as in sidewalks and fortifications; and 
takes the place of natural cement whenever the cost of the work 
is not thereby increased. 

Natural-cement mortar is used in the construction of ordinary 
walls, sewers, foundations for roadways, etc., when portland is 
considered too expensive. 

Puzzolan-cement mortar is employed wherever natural is 
used, but should not be employed as a surface coat. 

Quick-setting mortars are employed in harbor work between 
high and low tide, and whenever the masonry must, shortly after 
it is laid, be exposed to the action of moving water or to frost. 

The strength of cement mortar increases with age; in the 
first year a i-to-3 portland will attain a tensile strength of over 


3 8 4 


CIYIL ENGINEERING. 


400 pounds, and a i-to-2 natural a strength of over 220 pounds. 
The rate of increase decreases with the age. 

Mortar may be made water-proof by adding f of a pound of 
pulverized alum to each cubic foot of sand, and j of a pound of 
soft soap to each gallon of water used. 

Cement Concrete. 

Cement concrete is an artificial stone in which gravel or 
broken stone are bound together by cement mortar; the particles 
united are called the aggregate , the uniting material is called 
the matrix. 

The aggregate, like the sand in mortar, is introduced to 
reduce the cost of the artificial stone without reducing its 
strength below the limits required in practice. Besides gravel 
and broken stone, broken bricks, cinder, slag, shells, etc., may 
all be used as aggregate. The character of the aggregate de¬ 
pends largely on the cost and the ultimate strength required. 
It should always be clean. 

If the concrete is made by hand, the aggregate is spread in an 
even layer, 8 to 12 inches deep, on a platform of boards and 
thoroughly moistened with water. The mortar, made on an 
adjoining platform as previously described, is spread over the 
aggregate and the whole mass is mixed with shovels or hoes 
until every stone is thoroughly covered with mortar. 

If mixed by machinery, the proper proportions of aggregate, 
cement, and sand are dropped from an elevated platform into 
a revolving mixer, into which the water may be introduced through 
the hollow axle. The amount of water and the number of revo¬ 
lutions necessary for complete mixing are determined by experi¬ 
ment. When the concrete is thoroughly mixed it is dropped 
into cars or carts which are run beneath the mixer. This method 
is employed on large works. The mixer may be a cubical iron 
box revolving on a diagonal axle. 

Various forms of gravity mixers are also employed. 

The amount of mortar should be slightly in excess of the 
volume of voids in the aggregate. The volume of voids is deter¬ 
mined by filling a vessel, whose cubic contents are known, with 
aggregate, then adding water until the surface of the water 


NATURAL AND ARTIFICIAL STONE. 


3 8 5 


coincides with that of the aggregate. The volume of the water 
poured into the vessel is the volume of the voids in the material 
when dry and loose. The volume of voids will be less when the 
concrete is rammed in place. The volume of voids may vary 
from 35 to 50 per cent of the volume of the aggregates. 

The proportions by measure commonly employed in the 
manufacture of concrete are: natural cement 1, sand 2, aggregate 
4 to 5; portland cement 1, sand 3 to 4, aggregate 4 to 8. The 
proportions vary with the strength required. The most economi¬ 
cal proportions can be determined only by varying the proportions 
of sand and aggregate and studying the result. 

Use.—Concrete is employed in the construction of walls, 
arches, floors, sidewalks, foundation for roadway pavements, and 
is especially valuable in fortification work on account of its resist¬ 
ance to the penetration of large projectiles; in submarine work, 
where it may be laid when necessary without excluding the water; 
and in foundation work, where it is employed in leveling up for 
the lowest course of brick or stone work. 

Strength.—At the end of a year the crushing strength of 
natural-cement concrete is about 800 pounds, and of portland 
2000 pounds per square inch. The tensile strength of concrete 
cannot be accurately given; if carefully made, it is probably safe 
to say that its strength per square inch will be about one-tenth 
its compressive strength as given below. 

Experiments seem to indicate that the modulus of flexure of 
portland-cement concrete is about one and one-half times its 
tensile strength. 

Artificial Cement Stones.—Artificial cement stones are usu¬ 
ally carefully made cement mortar which is rendered compact 
by ramming or compressing the mortar in molds. By making 
suitable molds the stone may be given any desired form. Beton- 
Coignet, used in France, belongs to this class. 

Ransome stone is an artificial stone made by a different 
process. The mortar is made of sand, silicate of soda, and water, 
and is compressed in molds in the usual way. The stone is then 
immersed under pressure in a hot solution of calcium chloride; 
a chemical change now takes place resulting in the formation of 
calcium silicate, which forms an insoluble cement, and sodium 
chloride, which is removed by repeated washings. 


3 86 


CIVIL ENGINEERING. 


/ 


\ 


Asphalt and Coal-tar. 

Both asphalt and coal-tar are used with sand to make bitu¬ 
minous mortar and concrete. 

Asphalt is a natural bituminous substance which is found 
impregnating stone, as the Val de Travers and Seyssel asphalt; 
in the form of pitch containing more or less organic and inorganic 
impurities, as the Trinidad and Bermudez asphalt; and also in 
the liquid form, as some of the asphalts of this country. The 
purest deposits, like the Bermudez, contain over 90 per cent of 
pure bitumen. 

The Trinidad Lake asphalt, which has been more extensively 
used in this country than any other, contains in its natural state 
about 50 per cent of impurities and is prepared for use in the 
following manner: The crude asphalt is placed in large stills and 
there subjected to a temperature of about 300° F. The lighter 
impurities rise to the top and are skimmed off, while the earthy 
matter sinks to the bottom; this process is called refining. While 
still hot, the asphalt is mixed with a heavy petroleum oil, which 
softens it and converts it into asphalt cement. The hot cement is 
mixed with hot sand to make asphalt mortar, and with hot 
gravel or broken stone to make asphalt concrete. Both mortar 
and concrete must be put in place while still hot. The percent¬ 
age of asphalt in the mortar is about ten, which is about the 
percentage of asphalt in asphalt rocks. These are therefore 
converted into mortar by reducing them to powder and then 
subjecting them to heat. 

Asphalt mortar is used for roadway pavements; it is also used 
for roofing, but as it is liable to crack when subjected to changes 
of temperature, it is not a good material for this purpose. 

Pure asphalt, being impervious to water, is used in coating 
water-tanks and reservoirs and for covering roofing-felt. 

Asphalt concrete is also used in roadway pavements and in 
foundations for machinery when vibration is to be avoided. 

Coal-tar and pitch, although not so good as asphalt, are em¬ 
ployed for the same purposes. 

For further information consult: Merrill’s “Stones for Building and 
Decoration”; Sabin’s “Cement and Concrete”; Butler’s “Portland Ce¬ 
ment”; Taylor & Thompson’s “Concrete Plain and Reinforced”; “Profes¬ 
sional Papers, Corps of Engineers, U. S. Army, No. 28.” 


CHAPTER XXI. 


MASONRY. 

Masonry is the art of erecting structures in natural and arti¬ 
ficial stone; it is ordinarily treated under the heads Stone 
Masonry, Brick Masonry, and Concrete Masonry. 

Stone Masonry. 

Rankine’s Rules.—Rankine gives the following general rules 
to be followed in the construction of stone masonry: 

1. Build the masonry, as jar as possible, in a series oj courses, 
perpendicular , or as nearly perpendicular as possible, to the direc¬ 
tion oj the pressure they have to bear ; avoid all long joints paral¬ 
lel to that pressure. 

2. Use the largest stones jor the joundation course. 

3. Lay all stratified stones in such a manner that the princi¬ 
pal pressure which they must resist shall act perpendicular, or 
as nearly perpendicular as possible, to the planes oj stratification. 
This is called laying the stone on its natural bed, and is of pri¬ 
mary importance in strength and durability. 

4. Moisten the surface oj dry and porous stones before bedding 
them, in order that the mortar may not be dried out too fast and 
reduced to a powder by the stone absorbing its moisture. 

5. Fill every part oj every joint, and all spaces between the 
stones with mortar ,* taking care at the same time that such spaces 
are as small as possible. 

These rules being followed, the strength of the masonry will 
depend upon the character, size, and shape of the stone, upon 
the accuracy oj dressing, and upon the bond. 

Character, Size, and Shape.—As has been previously stated, 

the allowable unit stress on a stone depends upon its compo- 

387 


388 


CIVIL ENGINEERING. 


sition and texture, and the strength of the mortar upon the char¬ 
acter of the cement and the proportion of the ingredients. Fine 
uniform-grained granite is ordinarily specified when the unit 
pressure on the masonry is very great; it is laid with strong 
portland-cement mortar. Where great strength is required the 
stones are from 2 to 2\ feet thick; in jirst-class masonry they are 
from 1 to 2} feet thick; and in second-class masonry not less than 
§ of a foot thick. 

The width of the stones is from one and a fourth to twice their 
depth, and their length is from twice to three times their depth. 
The stones used in first-class masonry therefore weigh from 1 to 
4 tons. The width of the face of a stone is called its bed; the 
depth is called its build. 

Accuracy of Dressing.—In order that the thickness of the 
layer of mortar between the stones shall be a minimum, the 
surfaces of the stones must be carefully dressed. The more 
careful the dressing the more uniform will be the pressure 
transmitted from one stone to another. In first-class masonry 
it is therefore specified that the bedding-joints shall not ex¬ 
ceed \ to J inch in thickness; the vertical joints are dressed 
with equal care, but only to a depth of 12 inches from the 
surface. In second-class masoyiry less care is taken; the stones 
are so dressed that the joints shall not exceed J to j of an inch 
in thickness. 

Bond.—The term bond means the method of uniting the 
masonry so as to form a homogeneous mass without planes of 
weakness. In forming the bond the stones are divided into 
stretchers and headers. A stretcher is a stone whose longest 
dimension is parallel to the face or back of a masonry wall; a 
header is a stone whose longest dimension is perpendicular to 
the face or back of the wall. The bond is formed by making 
each course of both headers and stretchers, and by laying the 
stones in the successive courses so that there shall be no con¬ 
tinuous vertical joints. The wall is thus bound together both 
horizontally and vertically, and the pressure on each course of 
masonry is uniform throughout. The ordinary form of bond 
is shown in Fig. 118. 

In first-class masonry the headers extend entirely through 
the wall if it is less than 5 feet thick; if over 5 feet thick, they 


MASONRY. 


3 8 9 


must be at least 4 feet long, and extend into me wall 20 inches 
more than the adjacent stretchers. The distance between verti¬ 
cal joints in consecutive courses is at least 1 foot. In special 
cases two stretchers are allowed to one header. In second-class 
masonry the minimum length of headers in thick walls is 3J 
feet, and three stretchers are allowed to one header. In general 




ELEVATION 

Fig. ii8. 


SECTION 


the headers should constitute at least one-sixth the jace 0] the 
wall. 

When the above methods do not give the desired strength 
special bonds are made by the use of steel or iron dowels and 
cramps. A dowel is a pin or bar which fits in holes drilled or 
cut in adjacent stones of consecutive courses, and prevents the 
stones moving separately in 
a horizontal direction. A 
cramp is a Z, or channel¬ 
shaped iron strap, whose 
long arm is laid in a joint 
and whose short arms are 
vertical and are inserted in 

holes formed in the stones. ^ 

Fig. i 19. 

In masonry lighthouses 

on exposed sites, the stones of the same course are also notched 
into each other to prevent the jumping out of the exterior stones 
under wave action in a violent storm. Fig. 119 shows the method 
of doweling and notching employed by the Engineer Department 
in the construction of the Spectacle Reef Lighthouse, Lake 
Huron. 

According to the amount of dressing, building-stones are divided 
into three classes: unsquared or rubble stones; squared , hammered , 


















































































































































39° 


CIVIL ENGINEERING. 


or rough-dressed stones; and cut stones. The first class covers all 
quarry stones that are dressed simply by knocking off acute angles 
and excessive projections; the second class covers stones that 
are dressed with a face hammer or an axe until the stones are 
roughly squared and the vertical and bedding joints are roughly 
dressed; the face of the stone is usually rock- or quarry-jaced, 
pointed, or hammered; the third class covers all squared stones 
with smoothly dressed vertical and bedding joints; the face 
dressing may be any of those described under building-stones. 

According to the principles of construction masonry struc¬ 
tures are of three classes: simple walls, retaining- and reservoir- 
walls, and arches. 

Simple Walls.— Definitions. —The face, back, batter, and pro¬ 
file were described under Retaining-walls. 

Facing. —The stones that have one surface in the face of the 
wall. 

Backing. —The stones that have one surface in the back of 
the wall. 

Filling. —The masonry between the facing and the backing. 

Course. —A horizontal layer of masonry; usually one stone 
in depth. 

Coping. —A course of stone placed on top of a wall that is 
exposed to the weather, to bind the wall and protect its masonry. 

Footing. —The courses of masonry at its base, which project 
beyond the face and back and serve to increase the bearing area. 

Quoins. —The stones at the angle when two walls meet. 

Joints.— The surfaces of contact of the individual stones. 

Bedding-joints. —The joints, usually horizontal except in arches, 
through which pressures are transmitted. 

Builds. —The joints normal to the bedding-joints. 

Simple walls are intended to resist vertical forces only; the 
bedding-joints are therefore horizontal. 

The masonry in a wall may be rubble, squared stone , or ashlar. , 

Rubble masonry is that made of unsquared or rubble stoned 
and may be laid either dry or in mortar. It is uncoursed or ran¬ 
dom rubble when no attempt is made to lay it in horizontal courses; 
it is coursed rubble when it is laid in horizontal courses, although 
all the stones in a course are not of the same depth. In laying 
an uncoursed rubble wall, the large stones are bedded as well 


MASONR y. 


39 1 

as possible and then the interstices are filled up with small stone. 
In laying a coursed rubble wall the wall is constructed in layers 
a foot or more in thickness, depending upon the size of the stone. 
Dry rubble masonry is used principally for inclosure walls; rubble 
laid with mortar is used for inclosure walls and for walls of low 
buildings. Inclosure walls are usually capped by a coping which 
is made of stones of greater width than the wall. 

Squared or hammered stone masonry is made of squared or 
hammered stone and is laid in mortar. If the stones in each 
course are of the same depth so that the bedding-joints are con¬ 
tinuous, it is called range-work; if laid in courses not continu¬ 
ous throughout, it is called broken-range work; if no attempt is 
made to lay it in courses, it is called random work; Squared 
stone masonry is employed in the construction of basements 
and other walls in which appearance is not a governing considera¬ 
tion. At the corners of a building constructed of this masonry 
cut-stone quoins are frequently used to tie the walls together. 

Ashlar masonry is that made of cut stone; it is usually des¬ 
ignated by the face finish, as rock- or quarry-jace ashlar. The 
bedding-joints are ordinarily continuous; when not it is called 
broken ashlar. Ashlar is also called range, broken-range, and 
random ashlar, depending on the character of the bedding-joints. 

Dimension-stones are cut stones all of whose dimensions are 
specified. 

Combined ashlar and rubble walls arc constructed when 
it is desired to secure a pleasing appearance at a less cost than 
that of a full ashlar wall. The ashlar forms the facing, and coursed 
rubble the backing; the two are united by ashlar headers which 
extend through or well into the rubble, or they may be united 
by cramps. Brick masonry is also used to back ashlar. 

The pressure of beams, wall-plates, etc., on masonry should 
not exceed the following limits: 

Rubble masonry with cement mortar. 150 lbs. per square inch 

Portland cement concrete . 250 

First-class masonry, sandstone. 200 to 300 

“ “ limestone . 300 “ 500 “ “ “ 

“ “ granite .400 “ 800 “ “ “ “ 

Weight and Allowable Compressive Stress.—The weight of 
stone masonry is usually estimated at 160 pounds per cubic foot. 






392 


CIVIL ENGINEERING. 


The following have been recommended as allowable stresses 
in rubble masonry: * 

Rubble masonry, lime mortar. 5 tons per square foot 

“ “ natural-cement mortar... 6 “ “ “ “ 

“ “ portland-cement mortar.. 8 “ “ “ 

Coursed rubble, Portland-cement mortar.. . 10 “ “ “ 

For first-class stone masonry the Boston regulations allow 
the following: 

Sandstone. 30 tons per square foot 

Marble and limestone. 40 

Granite. 60 “ “ 

In ashlar faced work no allowance over the strength of the 
backing is made for ashlar less than 8 inches in thickness; if 8 
inches or more in thickness, the excess over 4 inches is allowed. 

Retaining- and Reservoir-walls.—The same principles which 
govern the construction of a simple wall also govern the con¬ 
struction of a retaining- or reservoir-wall. In accordance with 
Rankine’s first rule the bedding-joints of a retaining- and reser¬ 
voir-wall should not be horizontal, but should be as nearly as 
possible normal to the resultant pressure. This is rarely ever 
done in practice because of the extra expense and because of the 
small angle made by the resultant with the vertical. *To resist 
the shearing action of the horizontal component of the resultant, 
it is customary in high walls to make the filling between the back 
and face of the wall of uncoursed rubble of large stones. Such 
w^alls have no continuous horizontal joints. 

Arches.— Ovals. Next to circular, the oval arches of an odd 
number of centers are the ones most often constructed in engi¬ 
neering practice. Whatever be the span and rise, the tangents 
to these ovals are vertical at the springing-lines. 

To Construct an Oval of any Odd Number of Centers.— 
In Fig. 120 on the span AR describe the semicircle AIR, and 
divide it into the desired odd number of equal parts. Draw 
radii to the points of division. Assume the first center J, so 
that AJ is less than BC, the assumed rise. Draw DJ parallel 
to EB. Assume center K and draw OK parallel to FB. Assume 
center L and draw PL parallel to GB. From P draw PQ parallel 


* Proceedings Am. Soc. C. Engrs., Sept. 1904. 










MASONRY. 


393 



to the chord GH, and from C draw CQ parallel to the chord 
HI. From their intersec¬ 
tion draw QM parallel to 
HI. Then will J, K, L , 

M, and N be the centers of 
the circular arcs forming the 
oval arc ADPQC. 

The Basket-handled Oval. 

—This is considered the most 
graceful form of oval and is 
employed in the construction 
of ornamental arched bridges. 

In the semicircle Fig. 120 
inscribe a regular hexagon of 
which RS is one side. Draw 
IS, and parallel to it TU. 

Draw SB, and parallel to it 
UW. Then will V and W be 
the centers of the oval arc TUR. In the basket-handled oval 
each circular arc subtends an angle of 60 degrees. 

Centers.—Since, in a completed arch, each semi-arch is sup¬ 
ported by the pressure of the other transmitted through the keystone, 

the arch must be supported during 
construction until the keystone is 
inserted. The frame employed for 
this purpose is called a center. It is 
composed of a cylindrical platform 
of planks or boards upon which the 
masonry rests, and a series of wooden 
arched or trussed ribs which sup¬ 
port the platform. For small arches, 
such as window-caps, culverts, sewers, 
etc., a center like that shown in Fig. 
121, but without the intermediate struts, may be employed; for 
somewhat greater spans the struts must be inserted; for masonry 
bridge arches the platform is supported by a very strong truss 
somewhat similar to the truss constructed to support a curved 
roof. The truss must be strong enough to bear the weight of the 
arch without deformation of the curve of its platform. 



























394 


CIVIL ENGINEERING. 


The small center shown in the figure rests on folding wedges 
by means of which it may be slowly lowered when the keystone 
has been inserted and its mortar has set. A similiar device is em¬ 
ployed under the supports of heavy centers, or they may rest on 
pistons which are supported, in iron cylinders, upon beds of sand; 
when the center is to be removed, the sand is allowed to escape 
slowly. Centers of large arches are not removed for several 
months after the keystone is laid. 

In laying the masonry on the center, care is taken to carry 
up the two semi-arches simultaneously. It is sometimes neces¬ 
sary to load the crown of the center to prevent its rising when 
the semi-arches have only reached the haunches. 

Construction.—In a right arch the width of each voussoir, 
measured along the intrados, is the same; and the coursing- 
joints are normal to the intrados. Although the voussoirs may 
increase in depth from the crown to the springing-lines, each 
stone is a simple right prism and can be readily cut. Five of 
its faces must be accurately dressed. The arch is constructed 
in the same manner as a masonry wall; the coursing-joints are 
continuous, and heading-joints break their continuity at every 
coursing-joint. If the arch is backed with rubble masonry, as is 
usually the case between the haunches and springing-lines of 
large arches, some of the arch-stones are made long enough to 
bond the arch and its backing. 

If the arch is oblique, the construction is not so simple. Three 
methods of construction have been devised: the ribbed method , 
the logarithmic method , and the helical method. The first is ap¬ 
plied to arches of slight obliquity, and consists in making a num¬ 
ber of small overlapping right arches, each with the same rise 
and span as in Fig. 122. The soffit is therefore not a con¬ 
tinuous surface; but the coursing-joints are normal to the pres¬ 
sures, and the arch is easily constructed. 

In the logarithmic method the heading-joints are all parallel 
to the head of the arch, as in a right arch. The coursing-joints 
are warped surfaces generated by constructing lines on the soffit 
normal to the heading-joints, and using these lines as directrices, 
and lines normal to the soffit as generators. The projections of the 
directrices on a horizontal plane are logarithmic curves, and hence 
the name of the method. The soffit is continuous; the coursing- 


MASONRY. 


395 


joints are nearly normal to the pressures; but the arch-stones 
vary in size and are difficult to cut. 

In the helical method the intersections of the heading-joints 
and soffit are helices parallel to the helix which passes through 
the intersections of the springing-lines and crown with either 
face of the arch. The intersections of the coursing-joints with 
the soffit are helices perpendicular to the heading-joint helices. 
Both systems of joints are approximately perpendicular to the 
soffit. The soffit is continuous; with the exception of the stones 
in the faces and along the springing-lines, the arch-stones are 
all of the same size and are easily cut; the coursing-joints are, 
however, more oblique to the pressures than in either of the other 



methods. In the helical method cut stones must be used in 
the face and along the springing-lines, but bricks may be used 
in the rest of the soffit. 

Use.—Masonry arches of large span are employed princi¬ 
pally in bridge construction. The Cabin John bridge, on the 
line of the Washington aqueduct, constructed by Captain Meigs 
of the Corps of Engineers, has a span of 220 feet and was for 
many years the longest span stone arch. 

There are now several in Europe with longer spans. At Plauen, 
Saxony, there is a five-center oval stone arch which has a rise of 
60 feet and a span of 295 feet. 

Pointing Masonry.—The joints of stone masonry are pointed 
to prevent water from penetrating the joints, and, when freezing, 
from forcing out the mortar; and also to secure a pleasing appear¬ 
ance. Pointing consists in cleaning out the joints to a depth of 
about an inch, brushing them clean, moistening the masonry, 
and then filling the joints with a rich portland-cemcnt mortar 

























































































































































396 


CIVIL ENGINEERING. 


well rammed in. The surface of the pointing is given a smooth 
finish with a pointing-iron. In order to avoid bringing exces¬ 
sive pressure on the pointing and thus forcing it out of the joint, 
the operation 4 of pointing is the last work done on the masonry; 
it is therefore done after the mortar has had time to set and the 
masonry to settle under the action of its final weight. The point¬ 
ing should not project beyond the face of the wall. 

Settling of Masonry.—Unequal settling is avoided by com¬ 
pleting each course of masonry before beginning upon the succeed¬ 
ing one, by bonding the masonry well, and by making the joints 
narrow and uniform in width. 

Effect of Temperature.—In very hot weather mortar is 
liable to be injured by the too rapid evaporation of its water; 
this interferes with its normal hardening or setting. To prevent 
this the stone should be moistened before it is laid, and the masonry 
itself should be kept moist until the mortar hardens or sets. It 
is especially necessary to protect the top of an unfinished wall 
when the work is temporarily suspended. 

In very cold weather the hardening or setting of the mortar 
is retarded by the cold, and in freezing weather the mortar is 
also disintegrated by the freezing and expansion of its water. 
It may be used in freezing weather if its temperature can be kept 
above the freezing-point long enough to allow it to set with suffi¬ 
cient strength to resist the disruptive effect of the frost. This may 
be accomplished by using a quick-setting portland, by using hot 
water, by putting salt in the water used in making the mortar, 
by heating the stones, or by covering the masonry, as soon as laid, 
with canvas, straw, manure, etc. As workmen are careless, it 
is usually better, however, to suspend work while the tempera¬ 
ture is below the freezing-point. 

Brick Masonry. 

The general rules given by Rankine (page 387) apply to brick 
masonry where applicable. Bricks being more porous than stone, 
rule 4 is more important in brick than in stone masonry. 

Brick masonry may be laid with cement, cement and lime, or 
with lime mortar. The Boston building regulations require first- 
class or fire-proof buildings to be constructed with cement mortar; 
second-class buildings to be constructed with cement and lime 


MASONRY. 


397 


mortar, and allow only third-class buildings to be constructed 
with lime mortar. 

Bond. The bond in brickwork is secured, as in stone masonry, 
by the use of headers and stretchers, and by breaking vertical 
joints. If each course of masonry consists of alternate headers 
and stretchers, the bond is said to be Flemish. In laying masonry 
with this bond, the center of each header is in the same vertical 
line as the centers of the stretchers above and below it. If each 
course consists wholly of headers or wholly of stretcheVs, the bond 
is said to be English. In the English bond as commonly con¬ 
structed there are five to seven courses of stretchers to one of 
headers. The English bond is more easily laid than the Flemish, 
and its strength is easily varied by changing the ratio of the num¬ 
ber of header to stretcher courses; it presents, however, a less 
pleasing appearance. If the face of the wall is made of stretchers 
only, the bond is made by cutting off the corners of the bricks 
in the usual header-courses and inserting diagonal headers, or by 
cutting the stretchers of this course lengthwise, removing the 
rear half and inserting normal headers. This is called a run¬ 
ning or blind bond. Where the above bonds do not give suffi¬ 
cient strength, metal strips may be laid in the joints. 

Terra-cotta blocks, used as furring, are bended to the wall 
with clamps. Walls which meet at an angle should be bonded, 
at vertical intervals of io feet, with T- or L-shaped iron anchors 
which extend several feet along each wall. 

Walls.—The least thickness of a brick wall is 8 inches; this 
is the thickness of the walls of a narrow and low dwelling. The 
approximate rule for high walls is to make the first 25 feet, 
measured downwards from the roof, 12 inches thick in dwellings, 
and 16 inches thick in other buildings. The next 25 feet is made 
4 inches thicker, and this rate of increase is preserved to the 
ground-line. From the ground-line to the footing the increase 
is 4 inches for every 10 feet. The change in thickness is usually 
made at the floors. The pressure of wall-plates should not 
exceed 150 pounds per square inch for walls laid with cement. 

Weight and Allowable Pressure.—The weight of brick 
masonry is usually assumed to be 115 pounds per square foot. 

The following allowable pressures have been recommended:* 


* Proceedings Am. Soc. Civil Engrs., Sept. 1904. 



398 


CIVIL ENGINEERING. 


Common brick, lime mortar. 7 tons per square foot 

“ “ natural-cement mortar. .. . 8 “ 

“ “ portland-cement mortar... 10 “ 

Hand-burned brick, “ “ 12 “ “ 

Brick piers are bonded by building into them, at intervals 
of 2\ to 3 feet, flat bond-stones having the full dimensions of 
the pier. The allowable load on a pier whose height does not 
exceed twelve times its least dimension is about seven-eighths 
of the allowable load on a corresponding section of a wall if the 
wall has the same thickness as the pier. 

Retaining- and Reservoir-walls. —Brick masonry, being lighter 
than stone masonry, is not so good a material for walls which must 
sustain a horizontal pressure. If employed in the construction 
of low reservoir-walls, the wall is usually made impervious by 
coating it with asphalt or other water-proof material. Sylvester’s 
process has also been employed with success. This consists in 
applying with a brush a hot solution of three-quarters of a pound 
of soap dissolved in a gallon of water, and twenty-four hours 
later a solution of one-half pound of powdered alum dissolved 
in four gallons of water. This operation must be repeated 
several times. The soap and alum form an insoluble compound 
in the pores of the brick. Brick may also be rendered impervi¬ 
ous by applying, with a soft brush, finely ground cement moist¬ 
ened with water; several applications are necessary. 

The soap and alum process is also employed to prevent efflo¬ 
rescence on walls. 

Arches.—Brick is extensively used in the construction of 
arches for window-cappings, fire-proof floors, sewers, culverts, etc. 

There are several ways of laying bricks in an arch. The 
bricks may all be laid as stretchers; the arch is then composed 
of concentric rings each 4 inches thick. This is called the ring- 
bond and is employed in the construction of all arches of small 
span. Where several rings are necessary, it lacks strength in a 
radial direction; this may be secured by laying metal strips in 
the radial joints. The arch may be laid with continuous radial 
joints, the bricks in each radial course being alternate headers 
and stretchers, and breaking joints parallel to the intrados. This 
is the radial bond. The objection to this method is that the radial 
joints are very wide at the extrados. This may be avoided by 



MASONRY. 


399 


using only wedge-shaped bricks. With ordinary bricks the joints 
are filled with pieces of slate or with flat stones. The two 
methods may be combined by dividing the face of the arch into 
sectors, and constructing alternate sectors of ring and radial 
bonds; the radial sectors are narrow and the ring sectors wide. 
This is the combined ring and radial bond. 

The strength of brick arches was tested in 1895 in Austria 
with the following results: 


FIRST TEST. 



Span, Feet. 

Rise, Inches. 

Thickness, 

Inches. 

Final Load, 
Pounds per 
Sq. Ft., on 
entire span. 

Special brick, lime mortar.. . . 
Ordinary bricks. 

4.42 

4.42 

1.58 

4.91 

3-94 

5-9 

1638 

1638 

applied with¬ 
out rupture 



SECOND TEST. 


1 

Span, Feet. 

Rise, Inches. 

Thickness, 

Inches. 

Breaking 
Load, 
Pounds per 
Sq. Ft., 
applied on 
half the span 

Special brick, lime mortar... . 
Ordinary brick, lime mortar. . 

8.85 

8.85 

5 • 3 1 
9.84 

3-94 

5 - 5 i 

491 

883 


The arches were leveled up over the haunches with concrete. 


THIRD TEST. 



Span, 

Feet. 

Rise, 

Feet. 

Width, 

Feet. 

Crown 

Thick¬ 

ness, 

Feet. 

Spring¬ 
ing Line. 
Thick¬ 
ness, 
Feet. 

Breaking 
Weight, 
Lin. Feet. 

\ 

Brick in cement mortar. .. 
(300 lbs. cement to 35 
cubic feet sand) 

74-5 

15:0 

6.65 

i -97 

3-6 

1.81 tons, 
or 6001 lbs. 
per sq. ft. 


The load was a uniform live load covering one-half the arch only. 


The arch in the third test supported nine-tenths the breaking 
load of a similar arch of cut limestone laid with the same kind of 
mortar. 











































400 


CIVIL ENGINEERING. 


Concrete Masonry. 

Walls.—Concrete walls are usually monolithic structures, 
molded in place, in forms made of planks and frames. The 
concrete is deposited in layers 6 or 8 inches in thickness, care 
being taken to disturb the mixture as little as possible in moving 
it from the mixing-platform to the forms. It is thoroughly com¬ 
pressed by means of a tamp or rammer weighing 30 to 35 pounds. 
Each layer must not only be put in place before the concrete has 
had time to set, but should if possible be put in place before the 
preceding layer has set. This will insure thorough bonding. If 
the preceding layer has set, it is swept clean, moistened, and 
covered with a coat of mortar to insure its bonding with the new 
layer. The tamping is continued until the surface of the con¬ 
crete is covered with a thin film of water, and the layer is com¬ 
pleted by filling up all the depressions in the surface with mor¬ 
tar. If, the work is executed in warm weather, the masonry must 
be moistened from time to time for days, and in large masses 
for weeks, to prevent the concrete becoming dry and friable be¬ 
fore it has thoroughly set. 

Long walls are often made in sections, to diminish the prob¬ 
ability of developing cracks, due to stresses caused by shrinkage 
in setting, and by expansion and contraction due to changes of 
temperature. Each section is of such length that each of its 
layers may be put in place before the preceding one has set. 

The surface of a concrete wall is made of mortar, an inch or 
more in thickness, which is deposited against the sides of the 
form; if this coat is put on after the concrete has set and the 
form is removed, it is liable to separate from the body of the wall. 

There is some difference of opinion as to whether the con¬ 
crete should be mixed with an excess of water or not; with ordi¬ 
nary workmen the results are usually better, and the work is 
cheaper if an excess of water is used or the concrete is wet. 

In the construction of breakwaters, piers, dikes, etc., exposed 
to wave action, the concrete is commonly made into heavy blocks, 
these are laid in the same manner as stone masonry if the water 
can be excluded from the site, but are thrown in at random 
if the water cannot be excluded. Blocks weighing 40 tons have 
been molded for this purpose. 


MASONRY. 


401 


Allowable Pressures.—The allowable pressures on concrete 
walls are: * 

Natural cement, 1:2:5 . 8 tons per square foot 

I I 21 4 . q • • << << it 

Portland “ 1:2:5. 15 “ “ “ “ 

it ( ( 

i:2:4. l6 “ “ “ “ 

Retaining- and Reservoir-walls—Monolithic retaining-walls 
are made as above described; if the wall is not continuous in a 
vertical direction, the joint between the sections may either be 
sloped to the rear in accordance with Rankine’s first rule, or it 
may be stepped. 

When concrete is employed in the construction of reservoir- 
walls only portland-cement concrete should be employed, and 
care must be taken to prevent leakage. To prevent cracks in the 
walls, they have been constructed in short sections with con¬ 
tinuous vertical joints. Leakage through the joints has been 
prevented by making a vertical well in the joint and filling it 
with liquid asphalt or clay puddling. The bottom of a reservoir 
is made water-tight by making it of two layers of small concrete 
squares, separated by a half-inch of rich mortar. The reser¬ 
voir may also be made water-tight by coating it with asphalt 
applied as a paint to the concrete or to a lining of canvas. 

Arches.—Concrete is employed in the construction of floor 
and bridge arches, culverts, sewers, etc. The arch-ring is usually 
a monolithic structure; if joints are necessary in its construction 
they should be radial ones. The strength of portland-cement 
concrete arches was tested in Austria in 1895 with the following 
results: 


Composition. 

Span, 

Feet. 

Rise, 

In. 

Thick¬ 

ness, 

In. 

Breaking Loads, 

Pounds per Square Foot. 

Portland-cement concrete. .. 

4.42 

4-52 

2-95 

1638 without rupture 

it it 

8.85 

9-05 

3-35 

1127 on half the arch 

it it 

13-3 

16. I 

3-94 

790 “ “ “ “ 

ft it 

• • • 

32.8 

39 0 

16.4 

512 “ “ “ “ cracked, 
2500 “ “ “ “ broke 

it it 

74-5 

15.0 

27.6 

2.24 tons per running foot over 
half the arch, which 
was 6.6 feet wide. 


* Proceedings Am. Soc. Civ. Engrs., Sept. 1904. 


















402 


CIVIL ENGINEERING. 


Reinforced Concrete. 

The allowable tensile and shearing stresses in simple concrete 
are too small to admit of its use in the construction of structural 
forms which are subjected to forces of flexure, as floor-beams and 
slabs, lintels, etc.; or in the construction of those subjected to 
tensile forces only, as water-tanks, etc. 

The allowable compressive stress in simple concrete, although 
about ten times its allowable tensile stress, is also too small to 
admit its use in the construction of compressive members of 
small cross-section which are subjected to heavy loads, as the 
columns of a building. 

The requisite strength, in tension and compression, can, 
however, be secured by imbedding steel or wrought-iron rods in the 
concrete; these will assist the concrete in bearing the longitudinal 
and shearing stresses to which it is subjected. Simple concrete 
thus strengthened is called reinjorced concrete. In this country 
the elements employed are good medium steel and carefully 
made portland-cement concrete. If the cross-section is small, 
no aggregate is employed in the concrete, which is then simply 
a cement mortar. Wrought iron may be employed in place of 
steel. 

The possibility of thus making composite structural forms 
arises from the following considerations. 

1. Cement mortar is a preservative of steel; it absorbs car¬ 
bonic acid and can be applied to the metal in the form of an 
impervious coating. 

2. Both concrete and steel have practically the same coeffi¬ 
cient of expansion for changes of temperature. 

3. The adhesion of cement mortar to the metal is sufficiently 
great to require a considerable force to separate them along the 
surfaces of contact 

The composite material is cheaper and resists atmospheric 
agencies and the effects of a conflagration better than steel alone; 
it is stronger than concrete alone, and is not so liable to be dis¬ 
figured or injured by cracks caused by contraction or expansion. 

Concrete.—The properties of concrete are dependent upon 
so many considerations—character and proportion of cement; 
character, size, and uniformity of grain and proportion of sand; 


MASONRY. 


403 


character, size, and uniformity of particles and proportion of 
aggregate; thoroughness of mixing, amount of water employed, 
and amount of compression—that a perfectly uniform product 
is impossible. It results from these considerations that the con¬ 
stants of concrete deduced by different experimenters are not 
the same, nor are their proposed formulas. 

The following table gives safe unit stresses for steel and 
portland-cement concrete, as recommended by good authorities. 



Steel. 

Concrete 1.3:6. 

Factor of safety. 

4 

5 

5 

Compression. 

Compressive fibers in a beam. 

15,000 

12,000 

400 

500 to 700 

Tension. 

15,000 

12,000 

50 

Bending. 

15,000 

12,000 

75 

Shear. 

12,000 

10,000 

50 to 75 

Coefficient of elasticity. 

29,000,000 

2,900,000 

Coefficient of expansion. 

0.0000064 

0.0000064 


Safe adhesion of cement mortar to steel 50 lbs. per square inch. 


Columns.—Columns such as are employed in buildings are 
either circular or square and have usually four longitudinal rods 
near the surface; an additional one may be 
placed in the axis. These rods are tied together 
at short intervals by radial wires or metal 
strips as shown in Fig. 123. The outer rods 
may, in addition, be surrounded by circum¬ 
ferential wires or by a wire helix. This latter 
prevents the bulging of the concrete. 

The strength of these columns is deduced on 
the hypothesis that the concrete and the steel 
rods will be shortened equally as the load increases, and that the 
columns will fail when the unit stress in the concrete becomes 
equal to its modulus of crushing. 

If w' = load borne by the concrete; 

d[' = area of cross-section of concrete; 

L = length of column; 

1 = amount of shortening due to IV'; 

E' = coefficient of elasticity of concrete. 

W' l 

Then -n = E r -7- = unit stress on the concrete. 

A' L 













































404 


CIVIL ENGINEERING. 


If W" = weight borne by the steel; 

A" = area of cross-section of the steel rods; 
L = length of the column; 

/ = shortening of steel rods; 

E" = coefficient of elasticity of steel. 

W" l 

Then -^77 = £"— = unit stress borne by the steel. 


From these equations we have, since j- is the same in both, 

W r W" IV' W" 

° r W':W"::E'A':E"A", or 


that is, the unit stress in the concrete is to the unit stress in the steel 
as the coefficient oj elasticity oj the concrete is to that oj the steel. 

Since the coefficient of elasticity of concrete is one-tenth that 
of steel, the unit stress in the concrete will also be one-tenth the 
unit stress in the steel. A square inch of steel will therefore offer 
the same resistance as ten square inches of concrete. 

For the breaking load of a column reinforced by longitudinal 
rods only, we have therefore 


W=s c '(A' + 10A") = 2qoo(A' + iod"), . . (554) 


in which W = the breaking load; 

5/ = the modulus of crushing of concrete = 2,000 pounds; 
A'=the area of cross-section of the concrete; 

^ 4 "=the area of cross-section of the reinforcement. 


The safe load is one-fifth of the breaking load. 

The formula above given has been shown by experiments 
to give safe results for columns in which the ratio of length to 
least dimension of cross-section does not exceed twenty. 

For a column which has, in addition to longitudinal rods, 
circumferential wires or a spiral imbedded near its circumference, 
an additional factor must be added to the above formula to 
represent the additional load which may be carried, due to this 
strengthening. From experiment it has been ascertained that if the 
distance between consecutive bands or turns of the spiral is less 
than one-sixth the diameter of the column, the circumferential wire 
will be as effective as 2.4 its area employed as a longitudinal rod. 

The breaking weight of such a column is therefore 






MASONRY. 


405 


W = s c '(A' + 10A" + 24T"') = 2ooo(A' +10 A" + 24^"'), (555) 

in which A" r = area of cross-section of the wire employed in mak¬ 
ing bands or spiral wrapping. 

The circumferential wire cannot replace the longitudinal 
rods, but must be employed with them. 

Rectangular Beams.—A beam may have its reinforcement 
on the tension side only, or on both the tension and compression 
sides of the neutral axis. There are many theories on the action 
of composite beams under flexure. One of the simplest is that 
which assumes that the coefficient of elasticity of concrete is con¬ 
stant, and places no reliance on the tensile strength of concrete. 
The errors introduced by these assumptions are on the side of 
safety if the ultimate strength of the beam is alone considered. 
The theory adopted is based on the revolution of the plane of cross- 
section about its neutral axis, as in the common theory of flexure. 

With Reinforcement of Tensile Fibers Only.—In the canti¬ 
lever beam whose cross-section is shown in Fig. 124, the con- 


A b A' Q 



Fig. 124. 


crete is assumed to resist compression only. The reinforcement 
is on the tensile side of the beam. 

Let ABCD and A'C' = cross-section of a cantilever; 

EF and £'=its neutral axis; 

GH = position of cross-section under action 
of a bending moment M ; 

U — reinforcement of tensile fibers; 
y = distance along EC measured from £'; 
y'=EC. 

y" = E'l or distance from neutral fiber to 
axis of reinforcement; 

s'" = unit longitudinal stress at units dis¬ 
tance from E ' along EC ; 
b = breadth of beam=A£; 











406 


CIVIL ENGINEERING. 


d =depth of beam = ^ 4 C; 

A" = area of reinforcement; 
s e = unit stress in the reinforcement A". 

Since the depth of the reinforcement is small, we may assume 
that the stress in the reinforcement is uniformly distributed over 
it; the unit stress being the stress at its axis, IJ. 

Since the tensile and the compressive stresses on the cross- 
section must be equal and the compressive stress varies uniformly 
from the neutral axis to the surface, we have 

/ ? s"'bV 2 

s"'byoy= =s e A". . . . (556) 


If s = s'"y' = stress at surface fiber CD, we have 

sby' 


= s e A". 


(557) 


From the equality of moments about the neutral axis, we have 


sby' 27' 

—X — + s.A”?'=M 
2 3 


or 


sby' 


+ s e A"y" =M. 


■ (558) 


For the elongation, IJ, of an elementary length of the axial fiber 
of the reinforcement, we have 


IJ=V= S,oL 


rr 5 


! 559 > 


and for the shortening C'H of the extreme fiber under compres¬ 
sion, we have 

sdL 


C'H—l" = . . 

E' 

From similar triangles we also have 

IJ : C'H : 

hence 


— • • a/ • 

?rr • • y • y » 


or 


±.± 

E'E' 

E"y"s my"s 


E" 

in which m = ~. 

E' 


E'y' 


y 


/ ’ 


• • • . (560) 


(5 61 ) 














MASONRY. 


407 


If we represent the distance IC' by d", we shall have 

y"=d" —y .(562) 

For any given beam all the quantities in equations (557), (558), 
(561), and (562) are known except 5, s e , y', and y". From these 
four equations we can determine the values of the unknowns. 

To determine a value for y' in known terms, the value of s e 
from equation (561) is substituted in (557) and (558). 

by ' 2 by ' 2 

~^— = my"A" or —- my"A"=o. . . (563) 


sby ' 2 msy" 2 A" s /by ' 3 

——l--—,- = M or M=—.( - + my" 2 A 

3 '/ /V 3 


n 


( 564 ) 


Substituting for y” in equation (563) its value from equation 
(562) we have 

by ' 2 

- 2 - - mA"(d"-y')= o, 


or 


y' 2j r 


2mA" y' 2mA" d" 


from which 


/= 


mA 


tr 


+ \ 


2md"A" m 2 A " 2 
-7-+ ■ 


b 2 


or 


10A" I 

/=— r-+\ 


2od"A" 100 A " 2 
-7-+- 


b 2 


( 565 ) 


This gives the distance of the neutral axis from the extreme 
fiber in compression. 

Substituting in (564) the value of my"A" from equation (563) 
we have 


M = ~. 

y \ 3 


5 /by ' 3 by 


f 2 


+ 


y 


n 


Substituting for y" its value from (562) 

In this equation — is the total compressive stress in the 

cross-section, and d "is its lever-arm with respect to the axis 

3 


























408 


CrSIL ENGINEERING. 


of the reinforcement. If the center of moments is taken at the 
point of application of the resultant compressive stress, and the 
total tensile stress in the reinforcement is represented by F, we 
shall have 

<*"-£)• • • • (567) 

In the design of beams and slabs it is necessary to assume 
values for d" and the safe values of 5 and s e . In beams d" is 
made nine-tenths d f and in slabs five-sixths d. The safe value of 
5 may be taken as 500 pounds, and that of 15,000 pounds for 
stationary loads, and 10,000 pounds for vibrating loads. 

Substituting these values in equations (565), (566), and (567) 
we may deduce the following table. To make interpolation simple, 
A" 

the values of in the first column may be plotted as the ab¬ 
scissas, and those of the second column as the ordinates of a 

A" 

curve which will give the value of / for any value of be¬ 
tween 0.004 an d 0.050. 


M~F[d"--)=s,A 


rr l 


A " 
bd 

y 

d" 

3 

K-r) 


Equa¬ 

tion 

(566) 

5=500. 

M 
bd 2 

Equa¬ 

tion 

(567) 

15.000. 

Equa¬ 

tion 

(567) 

10,000. 

0.004 

0.23c? 

0.9 d 

0.82 d 

0. i886g? 2 

0.00328 bd 2 

47.2 

49.2 

32.8 

0.005 

0.25 d 

0.9 d 

0.82 d 

0.2050 d 2 

0.00410 bd 2 

5 i -3 

61.5 

41.O 

0.006 

0.27 d 

0.9 d 

0.81 d 

0.2187 d 2 

0.00486 bd 2 

54-7 

72.9 

48.6 

0.008 

0.3 id 

0.9 d 

0.80 d 

0.2480c? 2 

0 .00640 bd 2 

62.0 

96.O 

64.O 

O.OIO 

0.33d 

0.9 d 

0.79 d 

0.2607 c? 2 

0.007906c? 2 

65.2 

118.5 

79.O 

0.020 

0.43c? 

0.9 d 

0.76 d 

0.3268c? 2 

0.015206c? 2 

81.7 

228.0 

152.O 

O.O3O 

0.50 d 

0.9 d 

o- 74^ 

0.3626c? 2 

0.022206c? 2 

90.7 

333 -o 

222.0 

O.O4O 

0-54 d 

0.9 d 

0.72 d 

0.3888c? 2 

0.028806c? 2 

97.2 

432.0 

288.0 

0.050 

0 - 57 ^ 

0.9 d 

0. 71c? 

0.4047c? 2 

0.035506c? 2 

101.2 

532.5 

355-0 


Since all the values of the seventh column are less than those 
in the eighth, those in the seventh column determine the allowable 

M , , , 

values of for steady loads. For vibrating loads the values 

A" 

must be taken from the ninth column if is less than 0.007 
























MASONRY . 


409 


and from the seventh column if A"/bd is more than 0.007. 
By plotting the values oiA"/bd as abscissas and those of M/bd 2 
as ordinates, a curve may be constructed from which the value 
of M/bd 2 for any value of A"/bd between 0.004 and 0.050. 

If in the third column 0.83d is substituted for 0.9 d a table 
may be constructed for slabs. 


Problem .—What must be the width of a reinforced beam 12 
inches deep to safely resist a bending moment of 100,000 pounds, 
if the area of the reinforcement is one hundredth of the area of 
cross section? 


Using the table, we have °’° 00 =65.2, hence 6=10.7 inches. 

& 6X144 0 ' 

If 6 is made equal to §d, as is considered by some engineers 

M 

the proper form of cross-section of a reinforced beam, , becomes 

bu 


3M 

2d 3 ' 


Area of Reinforcement. —Knowing the cost per square inch 
of the concrete and the steel, it is possible to select the amount 
of reinforcement which will make the cost of the beam a minimum. 
There is, however, another matter which must be considered, and 
that is, the concrete on the tensile side must not crack and expose 
the reinforcement. It is usually assumed that reinforced con¬ 
crete can be stretched —— without injuring the beam. This 

1000 

has not, however, been conclusively shown, although the cracks 
have been shown to be minute. Since the maximum unit stress 
allowed in the steel is only 15,000 pounds, the elongation produced 

will be only about —— or one-half the allowable elongation. 

J 2000 

With Double Reinforcement. —In the cantilever beam shown 
in Fig. 125 let there be an additional reinforcement KN on the 
compressive side of the beam. Assume the same nomenclature 
as before, and in addition let 

A nf = a ea of the compressive reinforcement; 
y"' = the distance of its axis from the neutral axis; 
d"' = the d'stance of its axis from HC ; 
s c = its unit stress. 






4io 


CIVIL ENGINEERING. 


Then will equation (557) become 

S -^+s c A"’=sA 
2 

B 


// 


( 568 ) 


-- 

- !- 

C_ F 


l 




A' Q 


N 


Equation (558) becomes 

sby ' 2 


Fig. 125. 


s e A"y' + s c A'"y'" = M. . . . (569) 


3 

HC':KN:IJ::-^:^r,:^r n also HC':KN:IJ::y':y'":y"; 


hence 

Equation (561) becomes 


c s S 

. -jtt. v rr. ._. . _e_ 

> -y y • 


msy ,n 


,rr 


msy" y"s 

S e ~ yf ’ s e = ytt • • • • ( 57 °) 


Substituting the values of s e and s c from (570) and inserting them 
in equations (568) and (569), we have 


sby' msy'" A'" msy" A" 

+ ;-= 0 


or 


2 

by ' 2 


y y 

+m(A'"y'"-A"y")=o. . . 


(571) 


M = 


sby ' 2 msy'" 2 A'" msy " 2 A" 


y 


or 

But 


s /by ' 3 \ 

M =-7 + my '" 2 A'" + my" 2 A "J . 


y'" = y' — d!" and y" = d" —y'. 
Substituting these values in (571) we have 
by ' 2 

J —+m(A'" + A")y' = m(A'"d'"+A"d '). 


• • • 


(572) 

(573) 


2 


• (574) 



























MASONRY. 


4 *i 

Solving with respect to y' we have 

. HI 12 ffl m2 

y' = -j(A"' + A")± s J T (A”’d’" + A"d") + p- (A+ A"f. ( 57s ) 

This gives the position of the neutral axis if we substitute for 
m its value, io. 

By substituting the same values in equation (572), we have 

[ 7 / 3 

- 1 ^+mA'"(y l —d '") 2 + mA"(d" -y'f\ . (576) 

To test a given beam under a given load we must proceed as 
explained above. 

sby f 

If a = distance of resultant of — and sA m from the extreme 

2 c 

fiber in compression, we shall have 

M=(^+s c A"^J{d"-a) . (577) 

and M = s e A"(d"-a) .(578) 

A" A" 

If A'" is expressed in terms of A ", as — or —, and y' and a 

4 2 

are expressed in terms of y , and s c is made equal to s e , tables may 
be formed as for a beam with a single reinforcement. 

Some engineers treat a beam of double reinforcement as a 
plate girder of which the reinforcements are the flanges, and 
alone resist the longitudinal stresses. Under this hypothesis 
the stress on the compressive rod must not be great enough to 
cause excessive shortening and consequent crushing of the con¬ 
crete. 

Shear.—The vertical and horizontal shear in a reinforced 
beam is determined as explained for beams in general; the unit 
shear must never exceed its safe value, 50 pounds per square inch. 
The strength of a beam to resist vertical shear is often increased 
by inserting stirrups, which are imbedded in a vertical or inclined 
plane of cross section and support the lower reinforcement. 
These tension-stirrups act in a manner similar to the compressive 
stiffeners of a p’ate girder. 








412 


C1WL ENGINEERING . 


Adhesion.—To prevent the reinforcement from moving longi¬ 
tudinally in the concrete, the reinforcing bars are usually twisted, 
made with lugs or with some similar patented device. 

Methods of Reinforcing.—There are many different forms of 
reinforcement devised by engineers and inventors. The typical 
forms are perhaps the Monier, Melan, and Hennebique systems. 

The Monier system, which has been applied to floors, slabs, 
conduits, and arches, consists in imbedding in the concrete a 
network of wires. The wires of the network not only resist the 
tensile stresses in the structures reinforced, but also prevent 
cracks caused by internal stresses in the concrete. 

The Melan system, which has been applied to floors and 
arches, consists in spacing I beams, either straight or curved, at 
intervals equal to about ten times the depth of the beams and 
filling the space between them with concrete. In arched bridges 
of wide spans the beams are replaced by riveted girders or 
trusses. 

The Hennebique system is the type of the method applied to 
buildings. Foundation piles are made in the same manner as 
columns, except that the rods are brought together at the lower 
end, forming a point which is reinforced by straps and a cast-iron 
cap. Footings for columns are made of concrete in which are 
imbedded right-angled systems of steel rods with vertical stirrups. 
Walls are made of concrete with vertical rods near the interior and 
the exterior, and horizontal rods at intervals to strengthen the 
wall and prevent cracks. Columns and floor beams are made as 
heretofore described; the beams have straight reinforcing rods 
near the bottom, and also curved rods which are near the top 
surface at the supports and near the bottom at the middle points. 
The floor slabs connecting the beams have two sets of parallel 
rods which are placed perpendicular to each other. 

The other systems of construction differ from the above in 
the forms of the reinforcing rods and metal, and in the exact 
method of their introduction into the concrete. 

Uses.—Reinforced concrete is now employed not only in the 
construction of foundations, walls, floors, columns, and roofs 
of buildings, but also in the construction of bridges, retaining 
and reservoir walls, water-tanks, and conduits, sewers, and other 
engineering constructions. 


MASONRY. 


413 


Arches.—Since reinforced concrete readily lends itself to the 
construction of arches, the Austrian engineers made the following 
tests in connection with those on stone and brick arches hereto¬ 
fore described. 


Kind. 

Span, 

Feet. 

Rise, 

Inches. 

Thick¬ 

ness, 

Inches. 

Breaking Lead, Pounds 
Square Foot. 

Monier mortar and wire. . . 

8.85 

IO.23 

1 -95 

1638 without rupture. 

Melan mortar and I beams. 

8.85 

IO.23 

i -95 

1638 without rupture. 

Monier mortar and wire. . . 
Melan 3.15 I beams, 3^ ft. 

13-3 

16. I 

3-94 

872 over half the span. 

center to center. 

Monier mortar and wire, 

I 3 * 1 

11 .4 

3 -i 5 

3120 cracked. 

3370 broke. 

arch 13 ft. 1^ ins. wide . . 

Monier mortar and wire, 

32.8 

39-4 

5-9 to 

7.87 

90 tons on half the arch 
made a crack. 

180 tons similarly placed 
caused failure. 

arch 6.6 ft. wide. 

74-5 

15.O 

19.8 to 
23-44 

3.09 tons per running foot 
on half the span 
caused failure. 


For further information consult Baker’s “Treatise on Masonry Con¬ 
struction;” Kidder’s “Building Construction,” Part I ; Marsh’s “Reinforced 
Concrete;” Buell and Hill’s “Reinforced Concrete;” Taylor and Thompson’s 
“Concrete, Plain and Reinforced.” 


i 
















CHAPTER XXII. 




FOUNDATIONS. 

General Principles.—In the treatment of foundations, it is 
customary to consider the material or soil upon which the struc¬ 
ture is to rest and the means employed to distribute the weight 
of the structure over this material. A foundation may be con¬ 
sidered satisfactory when the vertical movement or settling of 
the structure is uniform and hardly appreciable, and there is no 
horizontal movement of the structure on its base. 

If the soil under every part of the structure has the same 
bearing power, the settling will be uniform if the unit pressure 
is the same at every point of the base. This is secured by design¬ 
ing the base of every section of the structure so the resultant 
pressure shall pierce the base as closely as practicable to the center 
of figure. The amount of settling will then be hardly appreciable 
if the unit pressure on the base of the structure is but a small 
fraction of the unit bearing value of the soil. This is secured 
by enlarging the base of each foundation wall and pier; this 
enlargement is called the footing. 

If the soil under the structure is not uniform throughout, 
the unit pressure on the base at any point must be proportioned 
to the unit bearing power of the soil at the same point; this is 
effected by varying the dimensions of the footing. The greatest 
unit pressure must be where the bearing power is greatest and 
the converse. If a part of the structure rests upon rock and the 
remainder upon compressible soil, it will be practically impos¬ 
sible to so adjust the pressure and resistances as to prevent un¬ 
equal settling. The structure should if possible rest wholly on 
the rock or wholly on the compressible soil; if this is not pos¬ 
sible, a cushion of sand may be placed on the rock. 


414 


FOUNDATIONS. 


415 




If the center of pressure of any section of the base does not 
coincide with the center of figure, the pressure at that part of the 
base will not be uniformly distributed; and if the soil is uniform, 
the structure will tend to rotate about some line in its base. In 
the outer walls of buildings a slight tendency to rotate inwards 
is not objectionable, since it is resisted by the partition walls 
and floors and binds the parts together. 

The jactor of sajety of a foundation is the ratio of the unit 
bearing power to the unit pressure; it should be as large as prac¬ 
ticable if appreciable settling is to be avoided. 

The soil underneath a structure must be protected from 
all disturbing influences. On land the foundation walls of 
buildings should extend below the limits of frost, and in rivers 
the foundations of piers should be deep enough to be protected 
from the undermining action of the current. The extreme 
depth of the frost line depends upon the climate; in New York 
City the building regulations require the base of foundation 
walls to be at least jour feet below the surface of the soil. 

Lateral movement of a structure on the soil will be impos¬ 
sible if the horizontal component of the resultant pressure on its 
base is less than the friction produced by its vertical compo¬ 
nent, or if the resultant pressure makes with the normal to the 
surface of the soil a less angle than the angle of friction. 
Since this angle is about 33 degrees, structures may as a rule 
be safely constructed on the soil excavated to a horizontal 
plane. 

To avoid excessive excavation when the surface of the soil 
is much inclined, the base of the foundation wall may be 
a series of horizontal steps instead of a single horizontal sur¬ 
face. 

Classes and Bearing Powers of Soil.—Soils may be divided 
into two classes, firm and soft. Firm soils are those that will 
support the weight of any ordinary structure without undue 

settling; as rock, ordinary earth, dry sand, dry clay, etc. Soft 
soils are those that will not support the weight of an ordinary 
structure unless their bearing powers are increased by artificial 
means; as wet clay, wet sand, etc. 

The safe bearing power of rock is assumed to be about one- 
tenth of its crushing strength as determined by tests on small 


416 


CIVIL ENGINEERING. 


cubes. This gives safe bearing values varying from io to 120 
tons per square foot depending upon the character of the rock. 
It is rarely necessary to make the pressure on the base of a struc¬ 
ture greater than 10 tons per square foot. 

The following table of safe bearing values has been recom¬ 
mended.* 

Soft clay and wet sand. 1 ton per square foot 

Ordinary clay and dry sand mixed 

with clay. 2 tons “ “ “ 

Dry sand and dry clay. 3 “ “ “ “ 

Hard clay and firm, coarse sand. 4 “ “ “ 

Firm, coarse sand and gravel. 5 “ “ “ “ 

Testing Soils.—The values above given are only approxi¬ 

mate. When it is desired to measure the ultimate bearing 
power of a soil, a plate covering a square foot or a square 
yard of the material is loaded until a perceptible settling is ob¬ 
served, accompanied by a rising of the soil in its immediate vicinity. 
The saje bearing power is derived from the ultimate bearing power 
by dividing by a factor of safety whose value depends upon the 
uniformity of the soil. 

In testing rock it is desirable to know both the character and 
the extent of the formation. If the rock is near the surface its 
character is determined by uncovering a small area; its extent 
by sounding with a small iron rod or pipe. If the rock is at a 
considerable distance below the surface, it may be reached by 
any of the processes resorted to in driving artesian wells. The 
depth and character of the stratum may be determined by 
boring. 

In testing other soils the bearing power may be determined 
by loading a limited area after the surface has been excavated 
to the depth of the bed of the foundation. If the stratum upon 
which the building is to rest overlies a softer stratum, it is also 
necessary to determine the extent and thickness of the firm 
stratum. In Chicago where the surface soil overlies a very soft 
soil, the building regulations allow a load of 2 tons per square 
foot on sand and if tons on clay, when the bed is at least is feet 
thick; for thinner strata the loads must be reduced. 


* Proceedings Am. Soc. Civil Engrs., Sept. 1904. 









FOUNDATIONS. 


417 


CONSTRUCTING FOUNDATIONS ON LAND. 


In Firm Soils.—If the soil is rock and generally level, it is 
necessary only to remove all loose and decayed material under 
the piers and foundation walls and then fill the crevices and level 
the bed with a layer of cement concrete. The surface of this 
concrete becomes the bed of the foundation proper. If the rock 
is intersected by wide crevices, these may be bridged by masonry 
arches or by strong iron beams imbedded in concrete; if the 
surface is inclined, steps are cut in the solid rock for the bed of 
the foundation. 


If the soil is earth, sand, or clay, a trench is dug to a depth 
below the frost line and its bottom leveled to receive the footings 
of the foundation walls and piers. 

The jooting (Fig. 126) is a mass of 
masonry by means of which the base 
of a wall or pier is widened, so as to 




Concrete 


Fig. 126. 


decrease the unit pressure upon the / Brick ^ 
soil and at the same time move the l " ^ 
center of pressure farther from the 
edges. A footing is necessary whenever the safe bearing power 
of the soil is less than the unit pressure on the base of the wall 
or pier. A deep footing is usually stepped, since, with less 
masonry, it can be made as strong as a rectangular footing. 

The vertical thickness of the footing may be deduced by as¬ 
suming that the projecting part is a cantilever acted upon by 
a uniform upward pressure per square inch, equal to the unit 
downward pressure upon the entire base produced by the weight 
of the wall and its load. 

The width of the footing in feet at any cross-section is deter¬ 
mined by dividing the total pressure per running foot of the base 
of the wall, by the safe bearing power of the soil per square foot. 

According to standard building regulations the thickness of 
a concrete or stone footing should be at least one foot. Each 
stone must extend the full width of the wall. If the footing is 
stepped, each step of a concrete or stone footing is made one 
foot thick. The steps of a brick footing are either one or two 
bricks thick; if one brick thick the steps are 1^ inches wide, if 









418 


CIVIL ENGINEERING. 


two bricks thick the steps are 3 inches wide. A brick footing 
always rests on a bed of concrete. 

For the piers or columns of high buildings the footings are 
made either of several layers of steel I beams, rods, or bars laid 
at right angles to each other and imbedded in concrete,* or the 
bases of the piers are connected by strong inverted masonry 
arches which in turn rest on beds of concrete. In the New York 
regulations the arches must be 12 inches and the beds 18 inches 
thick. 

A foundation bed of sand or clay under buildings is pro¬ 
tected from the action of the water by the construction of suit¬ 
able drains; when necessary the foundation beds are protected 
from washing away by inclosing them in a suitable barrier' of 
sheet piling or other material. 

If springs are encountered, or water from any source runs 
into the foundation trench, the water is drained to a hole dug at 
some convenient place and then removed by pumping. 

Soft Soils.—The bearing power of the soft soils most nearly 
approaching in consistency to the firm soils may be increased 
sufficiently to bear the weight of an ordinary structure by increas¬ 
ing the area over which the weight is distributed. This may 
be done by making the footing wider, or by placing under the 
footing of the wall a deep cushion of compressed sand much 
wider than the footing itself. 

If the soil is marshy and the cushion is continually wet, a 
grillage made of several layers of beams in close contact may be 
employed instead of the sand. Temporary weights, such as the 
abutments of small military bridges, are often supported by 
mattresses of poles and brush tied together by wire. 

If the soil is too soft to admit of this construction, the trench 
under each foundation wall must be excavated until a layer of 
firm soil is reached, or the foundation must be strengthened by 
the use of piles. The soft soil should be removed by excavation 
and the foundation laid on the firm soil beneath whenever this 
can be done without great cost; this form of foundation is more 
reliable than a foundation on piles. 

Piles.—A pile is a column which is driven or forced into 


* Cambria Handbook, p. 299. 



FOUNDATIONS. 


419 

the soil. Piles are divided into several classes according to 
their purpose and the method of driving. 

Common Piles.—A common pile is a wooden pile driven into 
the earth by means of blows delivered on its head; it is intended 
to resist pressure either from above or from one side, or simply 
to compress the soil and increase its bearing power. The term 
bearing pile is also applied to a common pile which supports a 
load placed upon it. Common piles are trunks of trees, usually 
oak, elm, pine, or cypress. 

Short Piles.—If intended simply to compress the soil, common 
piles are from 8 to 15 feet long, 8 to 12 inches in diameter, and 
are driven close together over the bottom of the footing trench 
or over the area of the foundation. The weight of the structure 
rests on both the compressed soil and the piles. A modification 
' of this method is to make a‘ number of holes by driving and 
withdrawing a short pile; as soon as made each hole is filled 
with compressed sand. The structure then rests on the compressed 
soil and the pillars of sand. 

Long Piles.—If intended to support the weight of the 
structure or to resist great lateral pressure, the common piles 
are from 12 to 18 inches in diameter; their length depends upon 
their load and the character of the soil, and may be as great as 
100 feet. 

The upper or larger end of the pile is a plane surface normal 
to its axis, and is protected from brooming , or splitting, by sur¬ 
rounding it with a thick wrought-iron ring 1 to 3 inches wide. 
If the soil is soft, the lower end is also a plane surface; if the soil 
is hard, it is a conical surface, or a truncated cone with a base 
of from 4 to 6 inches in diameter; if the soil is very hard, the end 
is protected by a solid conical shoe which is strapped or bolted 
to the pile. In very soft soils the piles are also sometimes driven 
with the larger end downwards. The bark is usually removed 
before driving. 

Bearing-piles are also made of cast and wrought iron and of 
reinforced concrete. These are employed in places where wooden 
piles would not be durable or where the pile acts as a column 
without lateral support and great strength is required. 

Sheet Piles.— Common sheet piles are thick planks driven 
edge to edge in a vertical position so as to form a sheet or wall. 


420 


CIVIL ENGINEERING . 


Sheet piles are employed when it is desired to support the walls 
of a trench or make a temporary dam in a stream or pond not 
exceeding 20 feet in depth. 

If the material is soft, the lower ends of the sheet piles are so 
beveled that the resultant pressure of the earth will force each pile 
against the preceding one driven. The joints between the differ¬ 
ent piles may also be closed by driving the piles so as to overlap 
each other. In the Wakefield sheet-piling, each sheet consists of 
three planks so fastened together as to form a tongue and groove 
similar to flooring. Interlocking sheet-piling is also made of rolled 
or riveted steel plates of various shapes. 

Pile-driving Apparatus: Common Pile-driver.—The common 
pile-driver consists of two similar right-angled triangular frames 
of strong timber. The height of the driver is about 30 or 40 feet; 
its base is about half its height. The vertical posts are called the 
leaders or guides and are fastened to the inclined pieces of the 
triangular frame by horizontal braces. The two frames are placed 
side by side, about 2J to 3 feet apart, and are united by horizontal 
struts which connect the horizontal braces and beams and the 
two inclined beams; the pile-driver rests on rollers or is bolted 
to the deck of a scow or the floor of a car and is braced laterally. 

The pile is placed upright between the leaders and is held 
in place by temporary pieces which rest in brackets attached 
to the leaders. It is driven by a heavy mass of iron, called the 
hammer or ram , which slides on rails fastened to the inner surfaces 
of the leaders. 

The power employed in raising the hammer is applied to 
a rope which is fastened to the hammer and passes through a 
pulley attached to a cross-beam connecting the tops of the leaders. 
This power may be man, horse, or steam power. Man power is 
employed only in light work such as military bridging. The men 
pull on small ropes which are attached to the main hoisting-rope. 
This is called a ringing engine. 

When horse power is used, the hoisting-rope passes through 
a second pulley called a snatch-block, which is attached to the 
base of the pile-driver and thence to an ordinary capstan. 

If steam is employed, the hoisting-rope passes from the snatch- 
block to a hoisting-drum, which is held to its axle by a friction- 
clutch. When the clutch is tightened the drum rotates with the 


FOUNDATIONS. 


421 


axle and the hammer is raised; when the clutch is released the 
drum rotates in the opposite direction and the hammer is dropped. 

The rope may also be firmly attached to an axle whose rotation 
may be reversed, and to a nipper. The nipper is a weighted 
pair of tongs which slides between the leaders. The tongs are 
engaged in a ring in the ram when it is desired to raise the latter, 
and are disengaged when the hammer is to be dropped. The 
tongs may be disengaged by bearing against inclined plates 
near the top of the leaders or by means of a dropping device 
operated by a cord. 

Steam-hammer.—In the steam-hammer the blow is delivered 
by a heavy hammer attached to a vertical piston which has a 
stroke of about 3 feet. At the bottom of the driving-apparatus 
is a heavy ring which rests on the head of the pile and is held 
in place by the leaders and supports the steam-cylinder. The 
hammer is a heavy mass, weighing from ij to 2 tons, attached 
to the piston, and also held in place by the leaders; its head passes 
through the* base ring and strikes the pile. The steam-cylinder 
is at the top of the driving-apparatus, where it is supported by 
pillars attached to the base ring, and is connected to the boiler by 
a flexible tube. The driving-apparatus is attached to a 'rope 
which passes over a pulley at the top of the leaders; by means of 
the rope it may be raised or lowered. 

The steam-hammer drives piles more rapidly than the ordi¬ 
nary pile-driver; the blows, although of less energy, arc delivered 
with greater rapidity. 

The Water-jet.—If the soil is sand, silt, or mud without 
boulders or other obstructions a pile may be forced into the ground 
by placing a load on its top and then forcing a stream of water 
to its lower end. The water softens and stirs up the material 
so that the pile sinks under the load placed upon it. In sinking 
a wooden pile the stream passes through an iron pipe which is 
attached to the pile by means of staples. When the pile is in 
place the pipe is withdrawn. 

Piles which would be shattered by the hammer and reinforced 
concrete piles may be driven in this manner. 

For piers, wharves, lighthouses, etc., the piles sunk by this 
method are usually hollow cast- or wrought-iron piles about 8 to 
12 inches in diameter. The lower end of the pile terminates in 


422 


CIVIL ENGINEERING. 


a horizontal disk, about 3 to 4 feet in diameter, which greatly 
increases the bearing power of the pile. This is called a disk pile 
(I, Fig. 127). The water is forced into the pile through a cap and 
escapes through holes in the bottom of the disk. Iron disk piles are 
made in sections which are fastened together as the pile is driven. 

The Capstan.—If a common pile is converted into a screw 
by attaching near its bottom spiral blades, the pile may be screwed 
into the ground by fastening horizontal bars or a horizontal wheel 
near its top and thus converting it into a captsan. Such piles 
are called screw piles and are usually made with a shaft of solid 



wrought iron 3 to 5 inches thick and blades of cast iron 3 to 4 feet 
in diameter (II., Fig. 127). Screw piles may be forced through soils 
in which the water-jet cannot be employed and are used on shoals 
along the coast where the water is so rough that the ordinary pile- 
driver cannot be employed. Like the disk pile, the screw pile 
has a large bearing area to resist not only the downward pressure 
but also the lifting power of the waves. 

Load on Piles.—The resistance of a driven pile to further 
movement could be accurately determined at any time were the 
intensity, direction, and law of distribution of the earth pressure 
upon its sides and bottom known. As these cannot be definitely 
ascertained, all formulas for the safe load upon a pile must be 
theoretical, based upon hypotheses, or simply empirical formulas 
based upon observation. 

Theoretical Formulas.—Captain Sanders, Corps of Engineers, 
deduced a formula on the hypothesis that the work of the pile’s 
resistance developed by the last blow of the hammer is equal to the 
theoretic kinetic energy of the hammer at the instant of striking, 


Mv 2 

Rd= = Wh 

2 


or 


R = 


Wh 
d 


or 


j 


























FOUNDATIONS. 


423 


in which R = ultimate resistance of pile or its ultimate load, in 

pounds; 

M = mass of hammer; 

v — striking velocity of hammer, in feet per second; 

W = weight of hammer, in pounds; 
h = height of fall of hammer, in feet; 
d — penetration of pile at last blow, in feet. 

Recognizing that the value of R in this formula was too great 
because some of the kinetic energy of the hammer was expended 
in overcoming the friction of the hammer in the guides, in deform¬ 
ing the pile and hammer, in producing heat, etc., he proposed as 
the safe load of a pile one-eighth of R. His safe load was therefore 


R' = 


Wh 
8 d 


(579) 


To provide for the losses due to friction, etc., in the original for¬ 
mula it has also been proposed to increase the denominator of the 
theoretical formula by a constant determined from experiment. 

If the constant — is adopted, the formula becomes 

X 2 


R = 


Wh 

d+- 

12 


• ( 5 8 °) 


If h is left in feet and d is expressed in inches, this becomes 


Wh 12WJ1 
d 1 d T1 
12 12 


(581) 


If the factor of safety 6 is introduced, the formula becomes 


R = 


2WJ1 
d + i 


( 5 82 ) 


This is known as the Engineering News formula. 

Captain Mason, Corps of Engineers, deduced a formula based 
on the hypotheses— 

1. The work of the pile’s resistance during the last blow of 
the hammer is equal to the theoretical kinetic energy of the pile 
and hammer at the instant following the striking. 

2. The momentum of the pile and hammer at the same instant 











424 


CIVIL ENGINEERING. 


is equal to the momentum of the hammer at the instant of strik¬ 
ing. Let the nomenclature be as on page 423, and in addition let 
w and m= weight and mass of the pile; 

V = velocity of hammer and pile the instant after striking, 
in feet per second. Then 


7 MV 2 mV 2 (W + w)V 2 
Rd --b- 


and 

Mv = (M 4- m) V or 


TVv (W+w)V 


g 


g 


2 g 


hence V= 


(583) 


w 


W+w 


V. 


Substituting this value in the first equation, 

IW . , , W*h „ WVi 

Rd= W+^g’ ’ 8 ’ Rd= W+i> 01 R= w+m- 


For safety a factor of four is employed, or R' 


W 2 h 


( 584 ) 


4 (IV+w)d' 

If the weight of the pile is equal to that of the hammer, which 
is approximately true for long piles, this formula gives the same 
value as Captain Sanders’ formula. 

Other theoretical formulas have also been recommended 
with the losses due to friction and deformation introduced. 

The theoretical formulas above deduced give an approxi¬ 
mate result for the ultimate and safe loads of piles driven in firm 
soil immediately after the pile is driven, provided no great de¬ 
formation of the pile has taken place. The bearing power of 
such piles probably diminishes, however, with the lapse of time. 
The formulas cannot be applied to piles driven in very soft soils, 
in which the resistance is due to the friction on the sides alone. 
The safe loads of piles driven in very soft soils are greater than 
those given by the formulas. 

These facts led to the adoption of purely empirical rules 
for safe loading. The New York building regulations state 
simply that the piles shall not be less than 5 inches in diameter 
at the small end, nor shall any pile be loaded with more than 
20 short tons. A common rule is to limit the load for long piles 
to 200 pounds per square inch in very soft soil and 800 pounds 
per square inch in very firm soil. Probably the only satisfactory 
way of determining the ultimate and safe load is to drive several 
piles on the site selected and some time after they are driven 












FOUNDATIONS. 


425 


load them until they begin to sink. This will give the ultimate 
bearing power from which the safe bearing power may be ob¬ 
tained by the introduction of a factor of safety. 

The resistance of screw and disk piles is due principally to 
the resistance of the soil under the disk and screw. This is 
increased, however, by the resistance along the sides, which varies 
from 200 to 800 pounds per square foot of this area, depending 
upon the firmness of the soil. 

Method of Constructing Common Pile Foundations.—When 
the weight of the structure is to be borne by the piles, the piles 
are driven along the bottom of the foundation trench in rows 2\ 
to 3 feet apart; the piles in each row are similarly spaced. The 
tops are sawed off in a horizontal plane and are covered with a 
grillage or with a layer of concrete. 

The grillage is made by capping each row of piles with a 
heavy longitudinal beam, which is fastened to the piles by mor¬ 
tise and tenon joints or by drift-bolts. A plank platform may 
be spiked to the capping or one or more layers of beams placed 
at right angles to each other and in close contact may be laid be¬ 
tween the capping and the platform, each layer being fastened 
to the layers above and beneath. When a concrete bed is em¬ 
ployed, the earth about the tops of the piles is replaced by a bed 
of concrete which holds the piles in place and distributes the 
weight over the entire area as well as over the piles. 

If the soil to a considerable depth is very soft before the 
grillage is constructed, the piles are braced laterally by throw¬ 
ing loose stone in the spaces between them. This is usually 
done when the piles are driven into a firm stratum which is cov¬ 
ered by a thick layer of soft material. 

FOUNDATIONS UNDER WATER. 

The foundations of bridge piers and abutments, wharves, 
dock walls, breakwaters, etc., are constructed on sites covered 
with water. The difficulty of construction depends upon the 
depth of the water, the velocity of the current, the variation in 
depth of the water, wave action, and the care which must be taken 
to prevent settlement. 

There are two general methods of executing the work: in the 


426 CIVIL ENGINEERING. 

first the foundation is laid without draining the water from the 
site; in the second the water is drained from the site so that the 
work may be executed practically as in open air. Whatever 
the method of execution, if the structure is an important one, 
the soil must be previously examined by borings or soundings. 

# 

WITHOUT DRAINING THE WATER FROM THE SITE. 

Random Rock. —The simplest method of constructing 
foundations under water is to throw in layers of quarry stone 
until the stone reaches the surface of the water. The founda¬ 
tion is then leveled with a bed of concrete which becomes the 
base of the superstructure. This method is employed in the 
construction of breakwaters, jetties, etc. The foundation is 
rendered compact by the action of the waves and slight settlement 
is not objectionable. On exposed sites very large blocks of con¬ 
crete are preferred to natural stones. 

Cribs.—If the soil is firm, the foundations for light structures 
may be made by sinking common cribs. These are constructed 
of round or squared logs halved at the ends and firmly fastened 
together with bolts or iron cramps. The crib is divided into a 
number of pockets by strong cross-walls halved into the sides. 
Some of the pockets are floored near the bottom and others are 
left open. The crib is floated to the site, held in place by piles 
or anchors and sunk in place by filling the closed pockets with 
quarry stone. When the crib reaches the bottom, all the pockets, 
both open and closed, are filled with stone and the crib is thus 
securely anchored. The stone in the open pockets settles under 
those parts of the crib Which do not touch the bottom. Cribs 
require less stone than random rock foundations. If cribs are 
to be used for foundations of masonry structures the crib work 
should be terminated below the low-water line; for piers of tem¬ 
porary bridges the crib extends above the water level. 

Concrete.—If the bed is of exposed rock and the current is 
not strong, the foundation may be made of concrete. Since 
concrete dropped through water is injured by the mortar being 
washed from the stones, the following methods have been em¬ 
ployed to get it into place with as little injury as possible. The 
mortar is lowered to the bottom in open mesh bags which are 
carefully laid by divers; it is lowered in water-tight boxes with 


FOUNDATIONS. 


427 


doors in the bottom which can be opened by means of a cord 
when the box reaches the bottom; or it is passed through tubes 
which reach from the deck of a scow to the bottom. The con¬ 
crete is placed and rammed in layers. Since concrete is always 
more or less injured by being thus laid, and is not as uniform 
or compact as when laid on land, this method of construction is 
employed only when the other methods are too expensive. 

Piles.—If the bed of the river is of earth, and the water is 
not too deep, some form of pile foundation may be employed. 

If common piles are used as a foundation for a masonry 
structure, the piles are cut off and capped by a grillage all below 
the surface of the water. The piles may be cut off by a cross¬ 
cut saw fastened to a triangular frame as shown in Fig. 128 and 



worked by ropes, or with a circular saw fastened to a vertical 
shaft which is attached to a scow and rotated by proper machinery. 
The piles are strengthened laterally by the grillage and by broken 
stone or other material deposited between them. If the water is 
too deep to give the proper lateral support to the piles, the caisson 
method as described below is used. 

Iron disk and screw piles are made long enough to extend 
above the water and may be leveled by telescoping caps. They 
are stiffened laterally above water by struts and ties connecting 
each pile with at least three others. 

The Cushing Cylindrical Pier.—This is a concrete pillar resting 
on a pile foundation and surrounded by a cylinder of wrought 
or cast iron. Its diameter varies from 4 to 10 feet. To con¬ 
struct this pier, a cluster of piles covering as large an area as 
can be conveniently surrounded by the cylinder selected is driven 
well into the bed of the stream. The piles of the cluster are in 
close contact. The piles may be cut off just below the water line, 
although this is frequently omitted. The iron cylinder is lowered 













428 


CIVIL ENGINEERING . 


over the piles in sections 5 or 10 feet long, which are bolted or 
otherwise fastened together. The cylinder is forced by weights 
into the soil of the bottom as deep as is judged advisable, and 
built up to the desired height of the pier. The cylinder is then 
filled with concrete with or without first pumping out the water. 
The piers are usually constructed in pairs or groups and are united 
by braces. These piers are used in the construction of highway 
and short-span railway bridges. 

In firm soil and still water the cluster of piles may be omitted 
and the concrete made to rest on the soil itself; in soft soil the same 
method may be employed, but the diameter of the pier must be 
increased. 

The Common Caisson.—The common caisson is employed 
in laying foundations of a limited area, as bridge piers when the 
water does not exceed 20 to 25 feet in depth. 

The caisson (Fig. 129) is a flat bottom boat with vertical sides. 



The bottom of the caisson is a grillage made of two or more 
layers of heavy timbers placed in contact with each other and 
fastened together. Its sides consist of frames to which planks are 
spiked. The sides are held in place by blocks spiked to the 
bottom, and by iron rods which at the bottom hook into iron 
rings fastened to the grillage beyond the sides, and at the top pass 
through heavy cross-beams which are notched into the sides. 
When the caisson is completed it is thoroughly calked to make 
it water-tight. 

Each horizontal dimension of the caisson is from 4 to 6 feet 
greater than the pier, and its height is several feet greater than 
the depth of the water. The caisson is loaded with a few courses 























































































FOUNDATIONS. 


429 

of masonry to give it steadiness and is then floated to the site 
which has in the mean time been prepared to receive it. It is 
accurately put in position, and while the masonry is being laid 
is held in place by common piles, anchors, or cribs. When the 
caisson has sunk nearly to the bottom it is gradually filled through 
valves with sufficient water to sink it in place. If it does not rest 
evenly on the bottom the water is pumped out, the caisson floated, 
and the foundation of the caisson readjusted. When it rests 
satisfactorily on its foundation, the masonry is continued until it 
reaches the top of the caisson. The iron rods are then removed 
and the sides float to the top and are taken away. 

The bottom of the caisson may rest upon the soil itself if it 
is firm, especially in still water, where there is no danger of under¬ 
mining. The bed is prepared by dredging the soft material and 
then leveling it for the grillage. 

If there is danger of undermining, the caisson may rest on 
piles. These are driven as in preparing pile foundations on 
land, and are then cut off in a horizontal plane near the bed of 
the river. 

If the bottom is rock the bed may be leveled with concrete as 
explained heretofore. 

The Well or Open Dredging Process.—This term may be 
applied to the process by which a masonry or iron cylinder or a 
wooden crib is, by means of interior excavation, lowered to the 
stratum upon which the foundation is to rest. The method is 
applicable to foundations both on land and in water; in the former 
the excavation is made by hand and in the open air, in the latter 
the excavation is made by dredging machinery which works 
through the water. The ordinary form of wooden crib for deep 
foundations under water (Fig. 130) is rectangular in cross-section 
and has vertical or slightly battered walls. The lower section of 
the crib ^4^4, which may be called the dredging chamber , is 10 
to 20 feet high and is inclosed and divided by wedge-shaped walls 
whose lower edges form the cutting edge of the crib. The wedge- 
shaped walls are made of i2"Xi2" beams in close contact and 
their cutting edges are covered with iron plate. 

The upper part of the crib is divided by vertical walls into a 
number of rectangular cells; like the exterior walls, the walls of 
the cells are made of i2"Xi2" timber, laid either in close contact 


43 c 


CIVIL ENGINEERING. 


or with small intervals. The cells immediately above the wedges 
BB have their bottoms closed by them, and are used to hold 
the material by which the crib is sunk in place. All other cells 
CC are open at the bottom and are the wells through which the 
material is dredged. 

The iron cylinder is similar in construction, having a dredging 
chamber with a conical roof pierced by the dredging-wells. 



B 

B 


B 


B 


B 


B 

C 


B 


C 


B 














B 

B 

B 

B 

/ 

B 



PLAN 

Fig. 130. 


The walls of the cylinder and of the dredging-wells are of steel 
plates which are fastened to a connecting skeleton steel framework 
of the proper form. 

Simple iron cylinders to be filled with concrete, or masonry 
piers resting on a strong grillage, have been sunk in the same 
manner. In the piers circular dredging-wells must be left for 
dredging the material. 

The same process is used in sinking shafts on land. A circular 
ring of masonry resting on a wooden curb is slowly sunk into the 
earth by gradually removing the material from the interior and 
building up the ring. A depth of 250 feet has been reached in this 
manner. 











































































FOUNDATIONS. 


4 3 1 


Methods of Sinking.—The lower part of the crib is constructed 
on shore, and, when launched, it is floated into place and there 
held by piles, anchors, or common cribs. 

The closed cells are then gradually filled with gravel, broken 
stone, or concrete. When the crib rests on the bottom, the dredging- 
buckets are lowered and the chamber is dredged; as the resistance 
below is diminished and the load above is increased, it settles 
under its weight until its base finally reaches the desired stratum. 
All the cells are now filled with concrete laid under water, and the 
foundation is complete. 

The upper part of the crib may be constructed as a wooden 
coffer-dam with water-tight walls; when the crib is in place and 
filled with concrete, the water in the upper part may be pumped 
out and the masonry laid as in open air. 

Dredging Apparatus.—The character of the apparatus depends 
upon the depth of the water and the material to be excavated. 
The most common apparatus is the clam-shell bucket attached 
to the hoisting-chain of a crane. As its name suggests, it is a 
bucket made of two leaves each shaped like an ordinary scraper. 
When open their cutting-faces are vertical; when closed they form 
a semi-cylindrical bucket. A modification of this bucket, called the 
orange-peel bucket, is hemispherical in form and opens out into 
four sections. These buckets are made of steel and when dropped 
open penetrate the soil, due to their own weight. By a suitable 
device the buckets are closed while in the soil and the material 
is hoisted by the crane. 

Sand-pumps, buckets on endless chains, and other forms of 
dredging apparatus are also used. 

Cribs and cylinders can be sunk by this method to a great 
depth in a soil free from large logs and boulders. To ascertain 
whether the soil is free from obstructions, it is examined by 
soundings or borings sunk to the level of the proposed founda¬ 
tion. There is however always some danger that an unexpected 
obstacle will prevent further settlement. Up to a depth of about 
ioo feet the work may be aided by divers, but beyond that the 
work must be trusted more or less to chance. There is also 
some danger of the crib becoming wedged, due to the unequal 
settling of the sides. For these reasons the pneumatic caisson, 
as hereafter described, is preferred for depths less than ioo feet. 


43 2 


CIVIL ENGINEERING. 


Poughkeepsie Bridge.*— The cribs used in the construction 
of the cantilever bridge over the Hudson at Poughkeepsie were 
ioo feet long, 60 feet wide, and ioo to 115 feet high. The height 
was such that when in place the top of each crib was 20 feet 
below high water. The dredging-chamber was 20 feet high and 
was divided into two equal parts by a central longitudinal parti¬ 
tion. The wedge-shaped side walls of the chamber were 10 feet 
thick at the top and its central partition 16 feet thick. The 
dimensions of the upper cells were generally io'Xi2', and their 
walls were 2 feet thick. 

The deepest crib rests on a stratum of gravel 135 feet below 
high-or 127 below low-water level; the masonry pier, 2^'XSy', 
was constructed in a separate caisson which was lowered on the 
crib when the latter was in place and filled with concrete. The 
masonry itself begins at 7 feet below low water. 

The Hawksbury Bridge. —The piers of the Hawksbury bridge 
in New South Wales were constructed by means of large iron 
cylinders sunk in the same manner. The deepest pier rests 140 
feet below low-water level of the river; it is the greatest depth that 
has been reached in pier construction. The plan of each cylinder 
is a rectangle with semi-circular ends whose extreme dimensions 
are 24' X 52'. The dredging tubes were three in number, equally 
spaced along the longer axis and each 8 feet in diameter. The 
dredging-wells and sides of the cylinder were connected by vertical 
trusses. 

The Diving-suit. —In all the above methods the services of 
the diver are required. The diver wears a strong air-tight suit 
surmounted with a copper helmet which rests on his shoulders. 
The helmet is connected by a flexible tube with a small hand air- 
compressor above water. To enable him to see, the helmet has 
a glass face, and to keep him in an upright position the soles of his 
boots are heavily weighted. The diver is raised and lowered by 
means of a rope which passes under his arms. 

The pressure of the air in the helmet must be sufficient to 
withstand the pressure of the water on the flexible tube, or 15 
pounds above the normal for every 34J feet of submergence. 
The extreme depth to which a diver has gone is about 150 feet. 


* Trans. Am. v Soc. of Civil Engrs., vol. 18, pp. 199-216. 



FOUNDATIONS. 


433 


EXCLUDING THE WATER. 

Common Dams.—When the depth of the water above an 
impervious stratum does not exceed four feet and there is no 
current, a common earthen dam may be constructed around the 
site to be occupied by the foundation and the water excluded 
from the site by pumping. The top of the dam is two to three 
feet wide and the sides are allowed to take their natural slope 
under water. It cannot be employed in a current which washes 
away the material of the dam. 

Coffer-dam.—In water between four and twenty feet in depth 
with a good bottom, the coffer-dam may be employed. Its sides 
are sheet piling, which reduces the amount of material required 
in the dam proper and protects the material from the current. 

To construct a coffer-dam (Fig. 131), two parallel rows of 



common piles are driven around the site of the proposed structure. 

To strengthen them longitudinally and laterally, the piles 
of each row are connected by heavy horizontal beams called 
wales bolted to the piles above the water level, and the wales 
themselves are connected by rods or cross-beams. If the wales 
are between the rows of piles they will serve also as guides to the 
sheet piling; if on the opposite sides of the piles, as shown in the 
figure, additional horizontal pieces are bolted to the piles to serve 
as guides. If the water is deep, additional guides are placed 
between the water level and the bottom, which are held in place 
by battens spiked to the upper guides. 

The sheet piles are driven into the impervious soil and form 
a coffer , which holds the puddling or dam proDer. The sheet 
























































434 


CIVIL ENGINEERING. 


piles may be spiked to the guides or held in place by riband 
pieces as shown. 

The puddling should consist of material which is impervious 
to water and which will not be washed away if a leak is developed 
in the dam. A mixture of clay, sand, and gravel is the best 
material for the purpose; the finer material fills the interstices 
of the coarser, and the coarser material resists the movement of 
the water in case a leak is developed in the dam. Common 
loam, clay, and fine gravel are also employed as puddling mate¬ 
rial. The puddling is placed and tamped in such a manner that 
there shall be no distinct layers. 

Experience has demonstrated that a thickness of two feet of 
good clay puddling is sufficient to prevent the percolation of the 
water through a coffer-dam of the usual height. As a dam of 
that thickness has a small resisting moment and brings great 
stress on the common piles, the thickness of low dams is made 
equal to their height and that of high ones equal to half their 
height. 

The sheet piles must support the pressure of the puddling 
which is counteracted on the outside row, by the pressure of the 
water; the outer sheet piling may therefore be thinner than the 
inner. The thickness of the sheet piles depends upon the dis¬ 
tance between the guide pieces. The ordinary piling is two and 
one-half to three inches thick, but this must be increased if the 
dam is high and the distance between the guides is great. The 
piles must be driven sufficiently deep to prevent the puddling 
from escaping underneath them. 

The common piles must, with the weight of the dam, resist 
the tendency of the dam to overturn or slide, under the water 
pressure against the outside sheeting. This will determine the 
length of the piles and the distance between the piles in each 
row; the latter is usually from four to eight feet. 

The distance between the inner row of piles and the founda¬ 
tion is at least three feet if no inner excavation is necessary; if 
the soil must be excavated, this distance is increased by one and 
one-half times the depth of the excavation. When necessary 
the stress in the common piles may be relieved by inner braces 
which rest against the foundation itself or against adjacent sides 
of the dam. 


FOUNDATIONS. 


435 


The water is removed and the interior space kept dry by the 
use of common, centrifugal, or force-pumps, depending upon 
the area to be drained and the amount of leakage. To prevent 
interruption duplicate pumps are provided. 

Modifications of the Coffer-dam.—When the bottom of the 
river is rocky, it may be impossible to drive common piles to sup¬ 
port the dam. Under such conditions each row of common piles 
may be replaced by a row of iron rods inserted in holes drilled 
in the rock, or by cribs or box-shaped boats which are floated to 
the site and sunk by loading them with stones. The puddling 
is placed between the lines of sheet piling supported by the rods, 
cribs, or boats as in the case of the common coffer-dam. 

Rectangular cribs made of heavy logs, halved into each other 
at the corners and fastened together by bolts or iron clamps, 
are employed to protect the dam when it is exposed to floating 
ice in a swift current. The cribs are properly cross-braced and 
have a floor to hold the rock used to sink them. 

When the bottom is rocky and the water is deep, great 
difficulty is often experienced in closing the joint between the 
rock and the puddling and in draining the interior, because of 
the seams in the rock. To close these seams an apron of con¬ 
crete is often laid around the dam. 

Pneumatic Caisson.—The plenum pneumatic process consists 
in excluding the water from a limited area by means of com¬ 
pressed air. The apparatus first employed was the diving-belL 
This was a strong metal box open at the bottom which was low¬ 
ered to the bed of the river by means of cranes. The workmen 
entered the bell while it was above water and were lowered with 
it. The water was prevented from entering by the pressure 
of the air which was increased as the bell sank. The air was 
compressed by some form of air-pump and forced into the interior 
of the bell through a flexible tube. 

The pneumatic caisson is simply a diving-bell whose sides 
are prolonged so that they project above the surface of the water. 
The workmen now enter and leave the bell through doors made 
in its roof. To prevent the sudden escape of the compressed 
air the workmen first enter a small air-tight vestibule in which 
the air pressure can be increased or diminished at will. The 
caisson above the diving-bell is simply a common caisson or coffer- 


43 6 


CIVIL ENGINEERING. 


dam with water-tight walls within which the masonry of the 
foundation is begun. 

Construction. —The caisson (Fig. 132) is divided into an 
air-tight working-chamber and a water-tight coffer-dam extending 



Fig. 132. 


from the roof of the working-chamber to some distance above the 
water surface. It is provided with air-locks and vertical shafts 
for communication and the removal of material, and pipes for 
forcing compressed air and water into the chamber, removing 
sand and mud, and carrying the lighting wires. 

The working-chamber is usually 8 to 10 feet high and is sur¬ 
rounded by wedge-shaped walls with cutting edges similar to 
those of the crib and the cylinder heretofore described. The 
partition walls, if any, have flat bearing surfaces to support the 
weight of the roof and its load. The roof of the chamber must 
be air-tight and strong enough to support the pressures to which 
it is exposed. If made of wood, it consists of several layers of 
12"X12" beams laid in close contact; if of iron or steel, it is 
composed of trusses or girders connected by I beams and covered 
with plates. 

The coffer-dam may be made of wood properly calked to make 






















































































FOUNDATIONS. 


437 


it water-tight or it may be made of iron properly strengthened. 
To decrease the height of a wooden coffer-dam, a grillage of many 
layers of 12" X 12" timber is placed upon the roof of the working- 
chamber. In the grillage the logs may be placed in close contact 
or may be separated by intervals filled with concrete. 

The vertical shafts are usually of iron plate, and if at any 
time filled with compressed air, must be air-tight. Those for 
communication are about 3 \ to 4 feet in diameter and are provided 
with air-locks. Those for dropping concrete into the working- 
chamber are smaller and have air-tight doors at the top and 
bottom. 

The air-locks , or vestibules, are air-tight chambers which have 
doors for communication with the outer air and the working- 
chamber, and are provided with air-cocks and escape-valves for 
increasing or diminishing the air pressure. As the doors open 
towards the interior of the caisson, they are kept closed by 
the air pressure whenever there is a difference of pressure on 
the two sides of the wall in which they are made. An air-lock 
may be constructed at the top of the shaft, at the bottom, or 
at any intermediate position. The first is the best position for 
the workmen in case of accident, but the locks must be moved 
as the caisson is built up. The second is the most conve¬ 
nient, but also the most dangerous. An intermediate position 
is sometimes secured by making the shaft discontinuous; the 
upper part contains the air-lock at its base, the lower part is 
tangent to the upper and overlaps it to the height of the lock. 
A section of the shaft itself may be converted into an air-lock by 
inserting horizontal air-tight doors. 

The material at the bottom of the river through which it 
is necessary to lower the caisson in order to reach a firm stratum 
is usually mud, sand, or silt. These materials can be most 
readily removed by utilizing the compressed air in the chamber 
or by means of a sand- or mud-pump. 

If the air itself is used, the discharge-pipe is continued to 
the bottom of the working-chamber and a curved elbow in¬ 
serted. The short arm of the pipe is flexible and its end 
is partially inserted in a pool of water formed at the bottom. 
Bv opening a valve, the air rushes up the pipe and carries 
with it the water with which earth is mixed. The sand- 


438 CIVIL ENGINEERING. 

and mud-pumps are constructed on the principle shown in Fig. 133. 

Water under high pressure is forced into 
a chamber at the base of the discharge-pipe 
A, just below its junction with the suction- 
pipe B. As the water escapes between the 
pipes it creates a partial vacuum in pipe B, 
sucks up the mud or sand and water and 
discharges it over the coffer into the water 
or into scows. Large boulders may be 
broken up and removed through the air¬ 
locks, or may be stored near the roof of the chamber and finally 
buried in the concrete when the chamber is filled. 

Incandescent electric light is the best method of lighting 
the air-chamber, as ordinary lights burn rapidly in compressed 
air and vitiate it. 

Sinking the Caisson. —The pneumatic caisson is commenced 
on shore, launched, partially loaded with masonry, floated into 
position, anchored and lowered to the bottom much in the same 
manner as the crib and cylinder sunk by the well process. When 
it comes to rest, compressed air is forced into the chamber, and 
the workmen descend into it and begin to remove the material. 
When sufficient material has been removed they either leave the 
chamber or take refuge in the shaft while the air is allowed to 
escape gradually. The weight of the caisson then causes it to 
sink a few feet. If more weight is needed, concrete may be added 
to that already placed on the roof of the working-chamber. The 
difficulty is usually, however, the reverse, that is, to prevent the 
caisson from sinking too rapidly. This can be regulated only by 
carefully excavating under the partitions or piers of the working- 
chamber. 

This process is continued until the caisson finally reaches the 
stratum upon which it is to rest. If this is gravel or approximately 
level rock, it is leveled with concrete and the chamber filled with 
concrete. The concrete is lowered through the small shafts. 
The lower doors of these shafts are closed and supported, and the 
material poured in; the upper doors are then closed, the pressure 
regulated, and the lower doors opened. Water is poured into the 
tubes before and after the concrete to prevent the premature setting 
of the concrete due to the heat of the compressed air. 













FOUNDATIONS. 


439 


If the bed is an inclined stratum of rock, it may be blasted 
to a horizontal surface, blasted into steps, or, if the inclination is 
not great, the foundation may be prevented from sliding by steel 
rods let into the rock. The caisson is stopped as soon as one of 
its edges rests upon any part of the rock, and if the foundation is 
built upon steps or an inclined surface, it is leveled with concrete 
laid in water. 

As the pressure in the working-chamber increases fifteen 
pounds for every 34J feet of submergence of the cutting edge, the 
depth at which this method can be utilized depends upon the 
amount of pressure which workmen can stand. This is about 
forty-five pounds above the normal, which makes the limiting 
depth about 100 feet. 

A combination of the pneumatic caisson and the open dredging 
caisson has been devised and utilized for very deep foundations. 

The Caissons of the New York Suspension Bridges. —The 
caisson of the New York pier of the Brooklyn Bridge is con¬ 
structed of timber and is 172 feet long, 102 feet wide, and 32 feet 
high; its sides have a batter of 10/1. The working-chamber is 
about 10 feet high, its side walls have a slope of 45 degrees, and 
its roof is a solid wooden grillage 22 feet thick. During construc¬ 
tion the roof was supported by temporary partitions and piers. 
The caisson rests on rock at a depth of 78 feet below mean high 
tide; its working-chamber is filled with concrete. Among the 
novel features of the caisson was a central shaft 7J feet in diameter 
extending 2J feet below the cutting edges. This shaft was kept 
full of water and was utilized to remove large boulders. Besides 
ordinary calking the chamber was made air-tight by a continuous 
sheet of tin in the body of the roof, and was protected against fire 
by a metal lining. A detachable wooden dam was employed 
above the grillage. The caisson of the Brooklyn pier was similar 
in construction; as the top of the masonry was always above water, 
no dam was employed. 

The caissons of the Williamsburg Bridge, New York, are of 
timber, but are only 76 feet long and 60 feet wide. Two caissons 
were employed at each tower. The walls of the caisson and the 
roof of the working-chamber are of solid construction, but the 
greater part of the caisson is of an open crib work strengthened by 
several longitudinal and transversal Howe trusses. The roof of 


440 


CIVIL ENGINEERING. 


the working-chamber rests on the lower chords of the Howe trusses, 
and the platform for the masonry on the upper chords. The 
trusses themselves rest on the walls and the partitions of the 
working-chamber. Where necessary a dam with timber walls 
was employed above each caisson. When finally in place the 
caissons were filled with concrete. 

The Forth Bridge. —One of the iron caissons employed in the 
construction of the Forth Bridge, Scotland, is shown in Fig. 132. 
The base of the caisson is 70 feet and the top 60 feet in diameter. 
The caisson proper is filled with concrete and extends to low-tide 
level. The iron detachable dam extends from the low-tide level 
to some distance above high tide, the range of the tide being 18 
feet. On account of the exposed position of the piers, the sides 
were made of two concentric cylinders of iron plate connected by 
strong bracing. The roof of the working-chamber was supported 
by several riveted trusses and I beams. The caisson was provided 
with iron shafts each 3J feet in diameter, one for men and two 
for materials. Tubes for lowering concrete were inserted in the 
latter when the caisson was in place. The air-locks were at the 
top. One of the piers of this bridge rests on rock 96 feet below 
the level of high tide. This bridge is shown in Fig. 140. 

Foundations below Quicksand. 

None of the methods heretofore described is practicable when 
the bed of the foundation lies below a thick stratum of quicksand, 
or fine sand and water which runs easily. The Poetch method 
of overcoming this difficulty consists of substituting, for the wooden 
or masonry walls of the foundation-shaft as heretofore described, 
walls of frozen quicksand. 

To form the walls a number of iron pipes, closed at the bottom, 
are sunk or driven through the quicksand to the desired depth. 
These pipes are placed at small and regular intervals along some 
polygon surrounding the site of the shaft. Smaller pipes, open 
at the bottom, are inserted in the first set, and both sets are con¬ 
nected with a refrigeration apparatus so that the freezing mixture 
flows into the smaller and returns by the larger pipes. A cylinder 
of frozen soil is thus formed about each pipe, and in time these 
cylinders interlock and form a wall within which the shaft may 


FOUNDATIONS. 


441 


be safely excavated. Pipes open at the bottom may be first 
driven when the closed pipes cannot be easily forced into the 
soil; these are called pilot pipes and are large enough to contain 
the closed pipes. 

Another method which has been utilized for overcoming 
the difficulties of working in quicksand is the forcing of cement 
grout into the sand by means of force-pumps The cement in 
setting converts the sand into a solid mass. 

For further information consult Patton’s “Practical Treatise on Founda¬ 
tions,” Fowler’s “Ordinary Foundations,” and Wellington’s “Piles and Pile 
Driving.” 


CHAPTER XXIII. 


BRIDGES. 

Definitions.—A bridge is a structure erected over a water¬ 
course to connect lines of communication upon opposite banks. 
A viaduct is a similar structure designed to carry a line of com¬ 
munications above the surface of the ground. 

The essential parts of a bridge are the substructure, by 
means of which the roadway is elevated, and the superstructure, 
which carries the moving load and transmits its weight to the 
substructure. The substructure is composed of abutments and 
piers; the superstructure, of the floor and the floor-carriers. 

An abutment is one of the end supports of the superstructure. 

A pier is one of the intermediate supports of the super¬ 
structure. 

The floor is composed of. the flooring , over which the travel 
passes; the longitudinal joists , which support the flooring; and 
the floor-beams or cross-girders, which support the joists and 
transmit the weight of the floor to the floor-carriers. 

The floor-carriers are the plate girders, trusses, arches , or 
cables which support the floor and transmit its weight to the 
abutments and piers. 

A span is that portion of a bridge between the centers of 
adjacent piers and abutments. 

A skew bridge is one whose axis is oblique to the longer di¬ 
mension of its piers and abutments. 

Wooden Trestles.—The simplest form of support for a via¬ 
duct is the wooden trestle-bent (Fig. 134). It is employed where 
the viaduct is on firm soil. The bent is made up of a capsill , 
AB\ a groundsill or mudsill, CD) two vertical posts; two batter - 
posts , AC and BD) and two cross- or sway-braces, AD and BC. 
The sway-braces are omitted in bents less than 10 feet high. 

442 


BRIDGES. 


443 


The interior posts may also be inclined and make with the 
batter-posts an inverted W. In 
railway construction all the mem¬ 
bers except the sway-braces are 
i2"Xi2 /r timbers; the sway-braces 
are 3-inch planks. In highway 
viaducts lighter pieces are em¬ 
ployed. The height of the simple 
bent is limited to 25 feet. If bents 
of a greater height are required, 
they are made in the form of super¬ 
posed bents as shown in Fig. 135. 

The space between consecutive horizontals is called a story and 
does not ordinarily exceed 20 feet in height. 

In high bents the posts may be continuous and be made of 
single or spliced pieces, or the bents may be constructed sepa¬ 
rately and superposed. In 
the former type the stories 
are made by bolting timbers 
in pairs to the posts. In the 
second type the superposed 
sills may be in absolute con¬ 
tact, or they may be separated 
by the longitudinal bracing. 

Simple bents are placed 
at intervals of 12 to 14 feet; 
high bents, at intervals not 
exceeding 25 feet. They are connected horizontally by the stringers 
which connect adjacent caps, and support the roadway. To give 
additional longitudinal strength, the batter-posts of simple bents 
are connected by horizontal or diagonal bracing similar to the 
sway-bracing. In high viaducts or trestles the intermediate sills of 
the different bents are connected by horizontal beams and the 
batter-posts of each story by horizontal or diagonal bracing. 

The floors may also be strengthened by a diagonal system 
of bracing connecting the consecutive capsills. 

In constructing the bents, the posts are mortised or notched 
into the sills and the joints strengthened by drift-bolts, dowels, 
or plates; the sway-bracing is bolted or spiked to the posts. 














































444 


CIVIL ENGINEERING. 



Fig. 136. 


In a viaduct constructed on soft soil the common trestle- 
bents are replaced by pile-bents. The simple pile-bent consists 
of a capsill supported on three or more piles. The cap is usu¬ 
ally but a short distance above the surface; where high bents 
are required the pile-bent is covered by one or more stories of 
common trestle. 

Steel Trestles and Towers.—A steel trestle-bent has but 
two posts, which are usually inclined. The posts are built-up 

columns which rest on masonry bases; the 
capsill is a plate girder, and the groundsill is 
replaced by a laced member attached to the 
posts; the sway-bracing is made of steel ties. 
In low viaducts on city streets the bents con¬ 
sist of posts and caps only; the former are 
bolted to immovable bases. 

High steel viaducts consist of a series of 
steel towers connected by the superstructure. 
This requires less material than a series of trestle-bents, as no 
longitudinal bracing is required between towers. A trestle 
tower (Fig. 136) is formed of four vertical or inclined posts. 
Each face of the tower is divided into stories and braced like a 
simple trestle-bent. At intervals the diagonally opposite posts 
are connected by horizontal braces. If the longitudinal dimen¬ 
sion of the tower is much greater than its lateral dimension, an 
additional post is inserted on each side. 

Cribs.—Crib piers are constructed of logs or beams fast¬ 
ened together and weighted as previously explained. They 
are employed in the construction of hasty or unimportant 
bridges. If the current is strong, the cribs are so placed that 
they present the vertex of an angle to the current. 

Masonry Piers.—Masonry piers, extending at least to the 
height of flood-level, are employed in the construction of all 
important bridges. The plan of the pier is usually rectangular, 
with the longer dimension parallel to the current (Fig. 137). If 
the current is strong, the shorter faces of the pier, at least below 
the level of high water, are rounded or pointed to deflect the water 
and thus reduce both the pressure on the pier and the undermining 
action of the current. These deflectors are called starlings. If 
exposed to floating ice in a strong current, a portion of each 








BRIDGES. 


445 


up-stream starling is inclined at an angle of 45 degrees with 
the horizontal, so that the ice will be forced up the inclined sur¬ 
face until it breaks of its own weight. This prevents the forma¬ 
tion of an ice-gorge and the consequent pressure on the pier. 
The inclined surface is called an ice-breaker and extends from 
a few feet below low water to a few feet above high water. 

The longer dimension of the pier is fixed by the width of the 
superstructure; the shorter dimension is made sufficiently large 




to give an ample bearing surface for the superstructure and a 
large factor of stability to the pier itself. For spans over 100 
feet the piers are not usually less than 6 feet thick. 

The top of the pier is crowned by a heavy overhanging cop¬ 
ing, and the faces have a slight batter which does not ordinarily 
exceed 12/1. The dimensions of the base depend on the direc¬ 
tion and intensity of the resultant pressure and the bearing 
power of the soil. If the base is much larger than the top, off¬ 
sets with water-tables are made in the faces. 

Piers are usually constructed of rubble or concrete with a 

facing of quarry-faced ashlar. 

Approaches and Abutments—The approach to a bridge 
may be either a trestle or an embankment. If the former, the 
abutment is similar in construction to a pier; if the latter, the 
abutment is a combined pier and retaining-wall. It is designed 


























































446 


CIVIL ENGINEERING. 


to resist both the pressure of the embankment alone, and the 
pressure of the embankment with the weight of the superstructure. 

If the side slopes of the embankment need not be supported, 
the abutment is a straight retaining-wall with a rectangular 
face, whose length is determined by the width of the superstruc¬ 
ture. If the side slopes must be supported, walls triangular in 
elevation, called wings , are added in prolongation of the straight 
wall. Both the top and bottom surfaces of each wing are usually 
made in steps. If the base of the abutment is liable to be sub¬ 
merged at high water, the wings are bent back so that the con¬ 
traction of the waterway is gradual. The wings of the abut¬ 
ment are also made perpendicular to the face; this is called a 
U abutment. In the U abutment the wings become counter¬ 
forts, and if the embankment is not very wide, they relieve the 
abutment proper of much of the earth pressure on it. The length 
of the wings is about one and a half times the height of the abut¬ 
ment. A straight abutment may also be strengthened by a 
wide central counterfort; it then becomes a T abutment. The 
T abutment is employed in railway bridges. The width of the 
counterfort is that of the track, and its length one and a half 
times the height of the abutment. It relieves the abutment 
of much of the pressure due to a heavy engine about to move on 
the bridge. 

Abutments are usually constructed of masonry, but for tern- 
porary or unimportant bridges they may also be of wood. A 
wooden abutment is usually made of squared logs placed on 
top of each other and held in place by anchor-logs fastened to 
the face-logs, and to transverse logs buried in the embankment. 
The transverse logs must be beyond the plane of natural slope 
passing through the foot of the abutment face. If the abutment 
is U-shaped, the wings should if possible be connected by anchor- 
logs. 

The Floor.—If a simple highway bridge (Fig. 138) is taken 
as a model, the -flooring consists of yellow-pine or white-oak planks 
about three inches thick, which are laid transversely and se¬ 
curely spiked to the joists. The opening between planks is \ 
inch; for convenience in making repairs, a continuous longi¬ 
tudinal joint is made along the center line of all long bridges 
having a roadway 16 or more feet wide. The sidewalks are of 


BRIDGES. 


447 


2-inch planks, and are usually separated from the roadway by 
the roadway carriers; if not thus separated, they should be ele¬ 
vated a few inches above the roadway. If the roadway extends 
to the carriers, longitudinal beams, at least 6 inches high, called 
wheel-guards, are spiked to the floor to prevent the carriers from 
being injured by the hubs of the wheels. A hand-rail is also con¬ 
structed along each side of the flooring. The minimum clear 
width of roadway should be io feet for a single vehicle and 16 
feet for two vehicles abreast. 



The joists are of yellow pine or other wood, not less than 
3 inches thick; they are spaced about 2 feet center to center. 
Joists of .adjacent panels either overlap or abut against each 
other on the floor-beams. An air-space is left between their 
ends. If the spans are less than 25 feet, the joists rest directly 
upon the piers and abutments and become the roadway carriers; 
the joists may then be rectangular wooden beams or steel I beams. 

The floor-beams or cross-girders connect the roadway carriers 
and support the floor; rectangular wooden beams, singly or in 
pairs, rolled steel I beams, and plate girders may be employed 
for floor-beams. 

Metal Floors.—When the roadway is to be paved, the pave¬ 
ment is supported on a buckle-plate floor which is riveted to steel 






448 


CiyiL ENGINEERING. 


I-beam joists. A buckle-plate is a steel plate, about 4 or 5 feet 
square and & of an inch thick, which is so bent that if placed 
horizontally its central point is about 2 inches higher than its 
edges. Metal floors are also made of channel-shaped plates. 

In railway bridges of less than 15 feet span the cross-ties 
are usually fastened to longitudinal stringers which rest on the 
piers and abutments; in bridges of longer span these stringers 
rest on and are fastened to the floor-beams. In bridges with 
metal floors the ties are embedded in the ballast, which is sup¬ 
ported by the floor. For short spans reinforced concrete floors 
of the Melan type are also constructed. 

Camber.—The floor of a bridge is slightly arched in order 
that it may not become concave under the heaviest live load 
to which it may be subjected. The rise of the arch is called the 
camber of the bridge. 

Floor-carriers.—According to the method of supporting the 
floor and transmitting its weight to the piers and abutments, 
bridges are classified as plate-girder , truss , arch , and suspension 
bridges. 

Plate-girder Bridge.—A plate-girder bridge is composed of 
two or more plate girders to which are riveted the floor-beams. 
These beams rest either on the flanges or on brackets riveted to the 
web. The ends of the girder rest on steel bedplates which have 
sufficient area to distribute the weight and reduce the pressure 
on the masonry within safe limits. In all plate-girder and truss 
bridges, if the friction is not sufficient to resist the lateral pres¬ 
sure from the wind, the girders and trusses are so fastened to 
the abutments that lateral movement is impossible. To prevent 
longitudinal movement one end is anchored to the abutment; 
to allow longitudinal expansion the other end is supported on a 
smooth plate if the span is short, and on steel rollers if the span 
is long. Plate girders are most economical when utilized for 
spans of 20 to 100 feet. 

Truss-bridge.—A truss-bridge (Fig. 139) is composed of two 
or more parallel trusses to which the floor-beams are attached, 
usually at the panel points of one of the chords. Trusses are 
ordinarily made of wood, wood and steel, or of steel alone. The 
drawing is that of a light steel highway-bridge truss for a span 
of 100 feet. 


BRIDGES. 


449 



Steel- and wrought-iron trusses may be either riveted or 

pin-connected. The riveted truss is employed when great stiff¬ 

ness or freedom from vibration is desired. The ordinary types 
of trusses are the Warren truss or riveted girder for spans from 

20 to ioo feet, the Pratt truss with parallel chords for spans from 

20 to 200 feet, and the Baltimore or similar trusses for spans of 
200 to 8oo feet. 


Fig. 139. 

The flanges of the Warren truss are usually made of two 
angles with or without cover-plates, like those of a plate girder, 
or of two channels. If the span is short, the flanges are con¬ 
nected at their ends by narrow rectangular plates; between the 
plates the triangular bracing is composed of angles either singly 
or in pairs. The angles and plates are either riveted to the flanges 
or both angles and flanges to a connecting plate.' If the span 
is long, all the members of the Warren truss are laced members. 

In a pin-connected truss the compression chord is made 
of two channels, either rolled or built up, connected by a top 
horizontal cover-plate. The inclined end- or batter-posts of the 
type shown in Fig. 139 are of similar construction and are rigidly 
connected with the compression chord. The tension chord is 






















45° 


CIVIL ENGINEERING. 


made of eye-bars. To stiffen the truss near the abutments the 
use of laced members instead of eye-bars is recommended in the 
first two panels. The vertical struts are made of two laced or 
latticed channels; in short trusses they are riveted to the upper 
chords so as to secure greater stiffness. The diagonals are eye- 
bars; in light bridges the counters have turnbuckles so that they 
may be adjusted. The vertical tie near the abutment may be an 
eye-bar or, if stiffness is desired, a laced member. The floor- 
beams are either riveted to the verticals or are suspended from 
the pins by eye-bar connections; the former is preferable. The 
lateral bracing is composed of laced angles and eye-bars. 

In the riveted truss the general construction is the same, but 
the members are usually all laced or latticed pieces and are riveted 
together. 

The length of the truss is usually governed by local condi¬ 
tions; it is ordinarily assumed that in a bridge consisting of a 
number of spans the arrangement is economical, if the cost of 
substructure and superstructure is approximately the same. In 
medium spans the height of the truss is about £ to J the span; 
in very long trusses this ratio is reduced. 

Cantilever Truss.—Cantilever bridges are of two general 
types. In the first type, shown in Fig. 140, each pier supports 


k- -260- -680*--- -350 < - — 4- 680 '-*145^-690*-— •* 



r: . . ._CjX-/p: £> . _ --!-!- Cp~O jn 

• — a T —$ ——-— — ■ —e—&— — —®- 

Fig. 140. 

a double cantilever tower with long arms, and the ends of the 
arms of adjacent towers arc the supports for a simple or suspended 
truss which connects them. In the Forth Bridge, Scotland (Fig. 
14°), there are three cantilever towers each about 330 feet 
high. The central tower rests on a base 120 by 260 feet, and is 
1620 feet long between the ends of its arms; it is connected with 


1 












































BRIDGES. 


451 


the other towers by riveted trusses 350 feet long. The shore 
ends of the other towers rest on the abutments. The suspended 
trusses are 150 feet above high-water level. Each tower rests on 
four masonry piers; the distance between adjacent piers of con¬ 
secutive towers is 1710 feet center to center. The St. Lawrence 
cantilever bridge near Quebec has a central span of 1800 feet 
between piers, center to center. 

In the second type, shown in Fig. 141, a simple truss resting 
on two points of support has attached to its ends cantilevers 



Fig. 141. 

whose extremities form the supports for a simple suspended 
truss. The bridge shown in the figure is that over the Hudson 
River at Poughkeepsie. It has two simple-truss spans of 525 
feet and three cantilever spans of 548 feet. The bridge over 
the Mississippi River at Memphis of the same general type has 
a simple-truss span of 621 feet, a single-cantilever span of 621 
feet, and a double-cantilever span of 790 feet. 

Howe Truss.—The Howe truss is the type of bridge which 
is usually constructed when timber and simple forms of iron 
are employed. It is the principal truss employed for hasty work 
in military operations. The chords, the main diagonals, and 
the counters are made of timber, and the vertical braces of iron 
rods. The floor-beams rest on and are supported by one of 
the chords. The king-post type is employed when the span 
does not exceed 30 feet; the queen-post with counterstruts, 
when the span does not exceed 40 feet; and the type shown in 
Fig. 142 for longer spans. In long trusses the main diagonals 
are in pairs and inclose the counter-diagonals; to secure rapidity 
in construction all the diagonals are of the same size. The 
verticals are in pairs, and for the same reason are of uniform 
size. The chords are made of three beams of uniform size and 
of the same width as the diagonals, which are laid side by side 
and bolted together. The diagonals abut against hard-wood 






































45 2 


CiyiL ENGINEERING. 


blocks or iron castings which are notched into the chords; the 
diagonals are held in place on the blocks by dowels. Washers 
are placed under the heads and nuts of the rods to prevent them 
from crushing the chords. 



Arched Bridges. —Arched bridges are constructed of masonry, 
of reinforced concrete, and of steel alone. The arch is the type 
of bridge which is constructed whenever it is desired to secure 
a structure of fine architectural appearance; it is the only type 
that can be made wholly of masonry. 

Masonry Arches. —The masonry arch has been employed in 
bridge and viaduct construction since it was first developed by 
the Romans. It is preferable to all metal structures, since it is 
practically indestructible if well founded, and requires little 
expense for maintenance. It is, however, more' expensive than 
a metal bridge, and its employment is limited to shorter spans* 
Within limits as to span, arched bridges admit of great variation 
in construction. 

According to their spans all existing masonry arch bridges 
may be divided into three classes: those exceeding 200 feet, 
those between 100 and 200 feet, and those less than 100 feet. 
In the first class are less than ten, and in the second less than 
fifty; nearly all are therefore in the third class. When bridges 
over 100 feet are to be constructed, it is usually found cheaper 
to construct a series of arches than to construct a single-span 
bridge. 

Form of Intrados. —The full-center arch is easily constructed, 
since the voussoirs all have the same cross-section; it is very 
stable, since the resultant pressure at the springing-line makes 
a small angle with the vertical. Its field of usefulness is limited, 
however, by the fact that the ratio of span to rise must be con¬ 
stant. To employ the full-center arch the height of the road¬ 
way above the tops of the piers must be greater than one half 




























BRIDGES. 


453 


the span. The full-center arch is therefore limited to the con¬ 
struction of high masonry bridges or viaducts. When these 
are of great height and length, the bridge is made of high piers 
connected by a series of full-center arches, or of two or more tiers 
of arches. A viaduct of this latter type in Saxony is 1900 feet 
long, 264 feet high, and has four tiers of arches. 

The segmental arch is also easily constructed, and the ratio 
of span to rise admits of great variation. The horizontal com¬ 
ponent of the thrust on the abutment increases, and the area be¬ 
tween the intrados and the chord connecting the springing-lines 
decreases with the length of the radius of curvature. A seg¬ 
mental arch can therefore be adjusted to any span and rise, but 
for long spans requires very stable abutments; in the long-span 
segmental arches of the first and second classes the abutments 
are usually natural ledges of rock. When a bridge over a river 
consists of a series of segmental arches, their springing-lines 
must be high enough above the water surface to give the desired 
clearance for navigation and the requisite waterway for floods. 

The three-center ovals (page 393) and the elliptical arches par¬ 
take of some of the properties of the full-center and segmental 
arches, but are not so easily constructed. The ratio of the span 
to the rise admits of great variation, and the horizontal component 
of the thrust at the springing-lines is less than in a segmental 
arch of the same span and rise. These arches have a greater 
area between the intrados and the chord connecting the springing- 
lines, and a greater clearance at every point, than a segmental 
arch having the same rise and span, and present a more pleasing 
appearance. Elliptical and three-center oval arches are usually 
constructed when the springing-lines are low and it is desired 
to secure the maximum clearance and waterway. This type 
is employed for single arches of moderate span, and also for a scries 
of arches. 

In the oval of many centers (page 393) not only the ratio be¬ 
tween the span and rise admits of great variation, but the curva¬ 
ture may be changed as often as seems desirable between crown and 
springing-line. This permits of the construction of an arch of 
the elliptical type which is very flat near the crown. Its clear¬ 
ance and waterway are even greater than the three-center oval. 
Arches of wide span and slight rise are usually of this type. 


454 


CIVIL ENGINEERING. 


Spandrel. —The face-walls of the spandrel may be continu¬ 
ous, or they may be pierced by transversal arches which rest on the 
arch-ring and support the roadway (Fig. 143). The latter arch 
presents the more pleasing appearance and has a smaller load 
on the arch-ring; the dimensions of the arch-ring and the abut¬ 
ments may therefore be reduced. 

Capping. —To prevent the percolation of water through the 
arch-ring it is coated with an asphalt or other water-proof coat¬ 
ing. The drainage from this coating is received in a cross' 
drain at the lowest point cf the extrados, and conveyed through 
the walls by a suitable conductor. 

Materials. —Masonry arches are made of stone, brick, or 
concrete. Stone arches have been constructed with spans of 
about 300 feet, brick arches with spans of about 150 feet, and 
concrete arches with spans of about 140 feet. Concrete arches 
of long spans are usually of the segmental type with steel hinges 
at the crown and springing-lines. 

Luxemburg Bridge. —This bridge (Fig. 143), which spans a 
deep valley adjacent to the city of Luxemburg, Europe, shows the 



Fig. 143. 


present development of arch construction. It consists of three 
stone arches; the outer arches are full-center arches with spans of 
71 feet, and the middle one is an oval of 277 feet span. As the 
abutments of the main span are underneath the surface of the 
ground, the visible span is about 260 feet. The arch-ring is 
4.7 feet thick at the crown and 7 feet at the springing-lines. ' The 
spandrel arches are full-center arches of 18 feet span. To save 
expense, the bridge, which is 60 feet wide, is made of two arched 
ribs, each 20 feet wide, separated by an interval of 20 feet. The 
roadway bridges this opening. By constructing one rib at a time 
the same center was available for both ribs. 




























BRIDGES. 


455 


Arches of Reinforced Concrete.—Arches of concrete rein¬ 
forced by steel are of modern construction. Those of the Melan 
type are made by imbedding parallel steel ribs in concrete. The 
ribs are placed about 3 feet center to center, are curved to the 
form of the arch, and extend well into the abutments. In small 
arches they are made of I beams, and in large ones of riveted 
• girders of the Warren type. In the Monier type wire nets are 
imbedded near the intrados and extrados. Modifications of 
these methods are to imbed near the intrados and extrados either 
flat or twisted bars spaced like the ribs of a Melan arch. Arches 
of the Melan type have been quite extensively employed in this 
country. An arched bridge at Topeka, Kansas, has one span 
of 125 feet, two of no feet, and two of 97 feet. This type of 
arch can be constructed more rapidly and cheaply than a masonry 
one and will probably wholly replace the masonry arch in highway- 
bridge construction. 

Steel Arches.—The roadway of a steel-arch bridge may be 
supported by two or more plate-girder ribs or by two or more 
open braced ribs. In a plate-girder arch the roadway is usually 
supported by a trestle which rests on the ribs. The open braced 
rib may be made by uniting a straight with a curved chord by 
riveted triangular bracing (Fig. 144). In this construction the 
floor-beams are usually supported at the panel points of the 
upper and straight chord. The open braced rib may also be 
made of two curved chords connected by riveted triangular 
bracing (Fig. 145), The roadway may then be supported on 
a trestle resting on the rib, or it may be suspended by rods 
attached to the rib. The arch presents a more pleasing appear¬ 
ance than the simple truss, and is more easily erected over 
deep depressions. 

Arched bridges are of three general classes: those with 
hinges at the crown and springing-lines, those with hinges at the 
springing-lines only, and those without hinges. The hinge is 
a heavy steel pin against which the arch or its segments abut, so 
that the arch can accommodate itself to expansions and con¬ 
tractions caused by variations of temperature. 

In the three-hinged arch the reactions at the crown and 
springing-lines can be determined with the same accuracy as 
those of the simple truss; for this reason it is prefererd by many 


456 


CIVIL ENGINEERING. 


engineers to the other types; its maximum deflections under 
heavy loads are, however, greater than in the other types. One 
of the most notable arches of this type is the railway bridge over 
the Viaur River in France (Fig. 144). The span of the arch 



Fig. 144. 

is 721 feet, its rise 176 feet, and the height of the roadway above 
the river 380 feet. The total length between abutments, which 
is 1344 feet, is made up of the central span, two cantilevers each 
226 feet, and two suspended trusses each 85 feet. The canti¬ 
levers and trusses reduce the crown thrust. The two ribs of which 
the bridge is made are 109 feet apart at the springing-lines and 
19 feet at the crown. 

In the two-hinged arch the reactions cannot be determined 
with as great accuracy as in the three-hinged one, but its maxi¬ 
mum deflections are less. It has also its advocates. Two of 
the most notable bridges of this type are those over the Niagara 
River below the falls. The double-track railway bridge which 
replaced ^he suspension bridge has a central arch connected 

with the abutments by 
two simple trusses. The 
arch has a span of 550 
feet, a rise of 114 feet; 
the railway is 225 feet 
above the river. Its gen¬ 
eral form is similar to the 
arch span in Fig. 144. The roadway bridge has a two-hinged 
arch of the type shown in Fig. 145, which has a span of 840 feet, 
rise 137 feet, and carries a 46-foot roadway 240 feet above the 
water level. 

The arch without hinges is not now employed for long spans; 
the bridge over the Mississippi at St. Louis, constructed in 1874, 
is one of the most notable examples. This bridge has one span 
of 520 feet and two spans of 502 feet each. 




















BRIDGES. 


457 


Suspension Bridges. —A suspension bridge (Fig. 146) is 
one in which the floor is suspended from two or more cables or 



chains which are stretched between the piers or abutments. 
This type is employed for very long spans, and for short spans 
over deep gorges when great stiffness is not essential. It is 
probably the only type which can be applied to spans exceed¬ 
ing 2000 feet; it has been proposed to utilize the suspension 
principle for bridges of 3000 feet span. For spans between 
1000 and 2000 feet the suspension bridge has been largely 
replaced by the cantilever. 

The essential parts of a suspension bridge are the main cables 
or chains , from which the floor is suspended; the towers , upon 
which these cables rest; the anchorages , which hold the ends 
of the main cables; the suspenders , by which the floor is attached 
to the main cables; the trusses , or other devices, by which the 
oscillations are checked; and the floor system itself. 

Main Cables. —The main cables are hemp or wire rope, 
wire or eye-bar cables. Hemp rope is used in the construc¬ 
tion of temporary bridges only; twisted wire rope is employed 
in the construction of bridges for short spans or light loads; 
wire and eye-bar cables are employed for long spans and heavy 
loads. 

Steel-wire cables are formed in place and consist of parallel 
wires. In the Williamsburg suspension bridge over the East 
river the wire used is No. 6, 0.19 inch in diameter. The 
cross-section of each cable shows 7696 wires, divided into 37 
strands of 208 wires each. The wire in each strand is so spliced 
that it forms a continuous thread; the strand is held by a pin 
at each anchorage. The strands are bound together in the 
form of a cylindrical cable. Each wire is protected from moist¬ 
ure by a coating of hot linseed-oil, and the whole cable is covered 























































45 8 


CIVIL ENGINEERING. 


by a sleeve of steel plate. The wire has a minimum tensile 
strength of 200,000 pounds per square inch. 

The eye-bar cable is preferred by some engineers to the wire 
cable because the tensile strength at any point of the cable can 
be proportioned to the stress at that point. The allowable unit 
stress of steel in the form of bars is, however, much less than in the 
form of wire; the weight of the cable must therefore be greatly 
increased. A notable bridge of this class is at Budapest, Hungary. 

The curve of the unloaded cable due to its own weight is by 
definition a catenary. If its weight, which is small compared 
with the uniformly distributed load, is neglected, the curve of 
the cable due to the latter load is a parabola. 

In Fig. 147, let the curve represent the cable as deflected by 
a load uniformly distributed along abc. 



Let b =the origin of coordinates; 
be = axis of X ; 
bd = axis of Y ; 
l =span; 

d = deflection of cable at middle point; 

W = weight per unit length of roadway; 

T' = the tension in cable at b ; 

T =the tension in cable at any other point; 
x and y — coordinates of any point. 

The forces acting on the cable to the left of any point (x, y ) 
between b and T are T' and Wx. Since the cable is at rest, 
their moments must be equal, or 

Wx 2 2 T'y 

2 —Tji or x ~ j,j 7 ~.(585) 

This is the equation of the curve assumed by the cable; from 
its form we know it is the equation of a parabola referred to 
rectangular coordinate axes through its vertex. 

To find the tension at b , in the equation of the curve substi- 









BRIDGES. 


459 


tute for x and y the coordinates of any known point, as that 
at the top of the tower; the resulting equation will be 

l 2 2 T'd , Wl 2 

4=~W~’ whence V=1 ~U' • • • (5 86 ) 


Since there is only tension in the cable, the value of T at 
any point between b and the tower, as x, must be equal to the 
resultant of the other forces acting on the cable between b and x. 
These forces are T f and Wx; hence T= V\V 2 x 2 + T' 2 . Since 
T increases as x 2 increases, the tension in the cable will be a mini¬ 
mum at b and a maximum at the tower. By making x=o, we 

find the minimum value of T is T'. By making x = - and sub- 

2 

stituting for H its value found above we have 


T = 


\ 


IT 2 / 2 IT 2 / 4 Wl 
4 6 \d 2 2 



(5 8 7) 


From the values of T' and T we see that if the span and 
loading remain constant, the tension in the cable may be de¬ 
creased by increasing its deflection. 

Towers.—The towers are constructed of masonry or steel. 
The height of the towers is fixed by the clearance which must 
be left for navigation under the bridge, and by the maximum 
tensile stress which is to be allowed in the cables. 

Each tower is subjected to a vertical force which is equal 
to the sum of the vertical components of the tensile stresses in 
the two branches of the cable at the top of the tower, and to 
a horizontal force which is equal to the difference between the 
horizontal components of the tensile stress in the two branches 
at the same point. The horizontal force reduces to zero when¬ 
ever the two branches of the cable make the same angle with 
the vertical through their point of intersection. As the hori¬ 
zontal force tends to overturn the tower, it is desirable to reduce 
its intensity to a minimum. This is effected by resting the cable 
on a saddle supported by rollers. This saddle has a slight longi¬ 
tudinal motion, which restores the equality between the angles 
whenever this equality is disturbed by the elongation or con¬ 
traction of the cable due to changes of temperature or to the 
oscillations produced by a moving load. The same effect may 
be produced by a rocking tower. 










460 


CIVIL ENGINEERING. 


Anchorages.—Each anchorage (Fig. 148) is usually a heavy 
mass of masonry which holds the anchor-plate to which the 

cable is attached. Its di¬ 
mensions must be sufficient 
to resist the ' vertical and 
horizontal components of the 
stress in the cable. The 
anchor-plate is laid in a 
horizontal position and is 
connected with the cable 
proper by a curved eye- 
bar chain. The pins of this 
chain are held in place by 
heavy blocks of stone imbedded in the masonry. A tunnel is 
left in the anchorage for these chains. The distance of the 
anchorages from the towers increases with the height of the 
tower, and decreases with the inclination of the cable at the 
anchorage; if the cable between the tower and anchorage is not 
loaded, the anchorage will be nearer the tower than if this part 
of the cable supports a portion of the roadway. The anchor¬ 
ages are also the abutments of the superstructure. 

Suspenders.—The suspenders are the vertical cables which 
connect the floor system and the cables. When two cables sup¬ 
port each end of the floor-beams, compensating devices are 
frequently introduced, so that, whatever be the relative positions 
of the two cables vertically, each will bear one-half the total 
load. 

Stiffening Devices.—If a concentrated load is moved over a 
suspension bridge in which each floor-beam is attached to a sus¬ 
pender, but the floor-beams are not rigidly connected, the load 
will cause each floor-beam to oscillate in a vertical plane. The 
amount of oscillation will be affected, the relative weight of the 
dead and live load, the dip of the cable, and the velocity of 
movement. 

Various methods have been devised for checking these oscil¬ 
lations. One of the most effective devices is the stiffening-truss. 
This is a vertical truss which is fastened to the floor-beams. The 
truss may be a single truss extending from tower to tower, or it 
may be made of two halves which have a joint at the middle 
























BRIDGES. 


461 


point. It is so designed that it can resist pressure acting verti¬ 
cally either upwards or downwards. A stiffening-truss is placed 
along each side of the roadway. When a concentrated load 
comes on a suspender the load is transferred from the cable to 
the truss as soon as the former yields at the point of juncture 
of suspender and cable; the truss distributes the load over the 
floor-beams and suspenders and thus over a long section of 
the cables. Wide bridges have a number of such trusses. 

A second method of checking the oscillations is to truss the 
cable itself. In this method the cable forms one of the chords of 
a vertical truss and is connected by suitable web bracing to a 
second chord. The auxiliary chord is designed to resist oscilla¬ 
tion stresses only. This method has been applied to bridges 
with eye-bar chains. 

A third method is to connect the roadway on opposite sides 
of each tower by cables running over the tower. A fourth is 
to attach guys to the cable at its points of maximum deflection 
under a moving load; these points are at the quarter points of 
the span. The guys should be in the same vertical plane as the 
cable and should be as nearly as possible perpendicular to it. A 
fifth is to attach each main cable to a small cable which lies in 
the same vertical plane as the main cable, but is below it. The 
small cable is attached at the foot of each tower and is convex 
upwards. A sixth is to anchor the roadway by cables fastened 
to rings in the rock below the bridge; this method may be em¬ 
ployed when the bridge spans a rocky gorge. 

Floor.—The floor system differs in no essential details from 
the floor system of truss-bridges. Its camber is greater than that 
of a truss-bridge. 

East River Bridges.—Two of the principal suspension 
bridges are the Brooklyn and the Williamsburg bridges over 
the East river, New York City. Both bridges are wire-cable 
bridges of about the same span, 1600 feet, but differ from each other 
in many details. The latter is shown in Fig. 146. The towers 
of the Brooklyn bridge are of stone 280 feet high, those of the 
Williamsburg bridge of steel 330 feet high. The clearance 
being the same, 135 feet, the deflection of the cable in the latter 
bridge is the greater. With the same maximum stress it will 
therefore carry a heavier load. In the former bridge the shore 


462 


CIWL ENGINEERING. 


spans are suspended from the cable; in the latter the cable be¬ 
tween the tower and anchorage is unloaded. In the former 
the stiffening-trusses are made in two halves jointed at the middle 
point of the bridge; in the latter the stiffening-trusses are con¬ 
tinuous between points outside of the towers; they form canti¬ 
levers for, the support of the shore trusses. 

Loads in Bridge Design. 

Highway Bridges.—Standard specifications require highway 
bridges to be designed to support the following dead and live loads. 

The dead load on a member is the weight of the member 
itself and that part of the weight of the bridge which is trans¬ 
mitted to it by the other members. Thus a joist supports not 
only its own weight but also the weight of an area of planking 
whose length is that of the joist and whose width is the distance 
between centers of joists. A floor-beam supports in addition to 
its own weight the weight of the flooring and joists in an area 
whose length is the distance between centers of floor-beams and 
whose width is the width of the flooring. This area is called a 
panel of the floor. 

Each member of the floor system is also designed to resist the 
greatest stress which can be produced in it either by the greatest 
uniformly distributed live load or by the greatest concentrated 
live load which will probably be placed on one of its panels. 
The concentrated load is placed in such a position that it will 
produce its greatest possible stress in the member considered. 
The uniformly distributed and concentrated loads are assumed to 
act separately or together; if they act simultaneously the uni¬ 
form load is confined to the area not covered by the concentrated 
load. The uniform load is that of a crowd of men. If con¬ 
fined in a contracted space inclosed on all sides, this may reach 
180 pounds per square foot; upon a bridge it would probably 
never exceed 150 pounds. This latter weight should be sub¬ 
stituted for that given below when conditions warrant it. The 
concentrated load is that of a steam road-roller or of a crowded 
electric street-car. 

Bridges are divided by Cooper* into four classes and are de- 


* Cooper’s Specifications for highway and railway bridges. 




BRIDGES. 


4^3 


signed for the loads given in the table. The first is for the 
heaviest city loads and the fourth for country highway travel. 


WEIGHT OF LIVE LOADS ON FLOOR. 


Uniform. 

1 st class: 100 lbs. per sq. ft. 
2nd “ 100 “ “ “ 

3d “ 100 “ “ “ 

4 th “ 80 “ “ “ 


Concentrated. 

24 tons uniformly distributed on two 12-ft. 
axles, 10-ft. centers, any part of roadway. 

24 tons uniformly distributed on two 12-ft. 
axles, 10-ft. centers, on railway tracks; 12 
tons on two 5-ft. axles, 10-ft. centers, any 
part, of roadway. 

18 tons uniformly distributed on two 12-ft. 
axles, 10-ft. centers, on railway track; 12 
tons on two 5-ft. axles, 10-ft. centers, any 
part of roadway. 

6 tons uniformly distributed on two 5-ft. axles, 
10-ft. centers, any part of roadway. 


Each member of the truss is designed to resist the greatest 
stress which will result from the combination of the dead and 
the following live loads. 


WEIGHT OF LIVE LOADS ON TRUSSES. 


Class of Bridge. 

Span. 


i st and 2d 

1 st and 2d 
1st 

1st 

2d 

2d 

3 d 

3 d 

3 d 

up to 100 feet 
over 200 * ‘ 

up to 100 ‘ ‘ 

over 200 ‘‘ 

up to 100 ‘ ‘ 

over 200 ‘ * 

up to 100 ‘ ‘ 

over 200 * ‘ 

1800 pounds per lin. foot on car tracks. 

1200 “ “ “ “ “ “ “ 

100 “ “ sq. “ outside car tracks. 

gQ c ( t t t t ft t t t t tt 

gQ < i t t t t it tt t t a 

1 i t t t t t t tt tt tt 

1200 “ “ lin. ‘ ‘ on car tracks. 

1000 “ “ “ “ “ “ “ 

Same as 2d class, outside car tracks. 
Interpolate for intermediate spans. 



The weight of the live load per linear foot decreases as the 
span increases because the probability of the bridge being sub¬ 
jected to a live load covering every linear foot decreases with 
the span. 

Merriman’s formula for the dead load per linear foot of a 
highway bridge is 

Weight in pounds = 140 +126 + 0.2U— 0.4/, . . (588) 

in which b = width of floor in feet, 
l = length of span in feet. 















464 


CIVIL ENGINEERING. 


Railroad Bridges.—The dead load is the weight of the track 
and the weight of the structure itself. 

Merriman’s formulas for the live load per linear foot of a rail¬ 
road bridge are 

Weight in pounds = 560+ 5.6/single track, . (589) 

“ “ “ = 1070+10.7/double track, . (592) 

in which l = length of span in feet. 

The live load is the weight, on all tracks, of trains as long 
as the bridge, composed of two of the heaviest engines on the 
road hauling a train of loaded cars. The train or trains are 
placed so as to produce the maximum stress on the member 
considered. The concentrated load is an assumed load which 
will produce as great stresses on the floor system as any real 
load. 

The maximum live loads for railroad bridges are, according 
to Cooper: 

Engine. 9,800 lbs. per lin. ft. 

Engine and tender. 8,125 “ “ “ “ 

Train load. 5,000 “ “ “ “ 

Concentrated load.120,000 “ on two axles 6 ft. apart. 

Impact.—In some specifications the uniformly distributed 
live load is increased by a certain percentage to provide for the 
increased stresses due to impact. This percentage for railroad 
bridges is derived from the following formula: 

P 4o,ooo 

P (L + 500)’.( 59 b 

in which P = the percentage, 

L = the length in feet of the uniform live load which 
produces the maximum stress in the member 
considered. 

Wind Stresses.—The pressure of the wind upon a bridge or 
viaduct tends to buckle the superstructure, move the structure ‘ 
on its supports, and overturn the piers, towers, and trestles. 
All these effects must be considered in bridge design. 

Intensity and Direction of Pressure.—The intensity and 
direction of wind pressure on any plane surface normal to its 







BRIDGES. 


465 


direction is approximately expressed by the formula 

p = o.oozj.i' 2 ,.(592) 

in which p = the pressure in pounds per square foot, 

v = the velocity of the wind in miles per hour. 

According to this formula the pressure of a wind w T hose 
velocity is 100 miles per hour is 40 pounds per square foot. 

As the maximum pressure on small surfaces is much greater 
than on large ones, it is customary in this country to assume 
a pressure of jo pounds per square foot on large surfaces, and 
to increase this to 45 or 50 pounds whenever it seems desirable 
to increase the factor of safety, as on unloaded railroad bridges. 

The wind pressure is greatest when its direction is normal 
to the vertical plane through the axis of the bridge, as it then 
acts on the largest surface and produces all the effects above 
described. If the wind blows parallel to the axis, its effect is 
felt principally by the supports. 

Wind-bracing.—Let the truss shown in Figs. 139 and 149, in 
which the floor rests on the lower chord panel points, be exposed 
to the pressure of the wind blowing normal to the planes of its 
trusses. The actual area exposed to pressure is the area of the 
vertical projection of its floor system plus twice the area of the 
vertical projection of one of the trusses. This area in square feet 
multiplied by 30 will give the total pressure in pounds upon the 
bridge. As the abutment ends of the truss are prevented from 
moving laterally, this pressure will buckle the trusses and bend 
the vertical posts unless these are braced to resist this pressure. 
To prevent deformations in the trusses under wind pressure it 
is customary to connect the vertical trusses by one or two hori¬ 
zontal ones as shown in Figs. 139 and 149. These horizontal 
trusses transmit the horizontal pressures to the supports in the 
same manner that the vertical trusses transmit the vertical 
loads. 

At the loaded chord the horizontal truss is made by simply 
connecting the floor-beams by diagonal ties (Figs. 139 and 149). 
The loaded chords of the vertical trusses thus become the ten¬ 
sion and compression chords of the w r ind truss and the floor-beams 
become its struts. At the unloaded chord horizontal struts as 
well as a system of diagonal ties are introduced between the 



466 


CIVIL ENGINEERING. 


chords of the vertical trusses. The effect of the wind pressure 
is therefore to increase the stress in the main tension chord on 
the leeward side and in the main compression chord on the 
windward side. In determining the stresses on the wind trusses 
the pressures are assumed to act horizontally at the panel points 
of the chords. 



In Figs. 139 and 149 the wind pressures are transmitted by 
the lower wind truss immediately to the abutments, but the upper 
truss transmits them only to the top of the end posts, from which 
points they must be transferred to the abutments by the shear¬ 
ing resistance of the end posts. To prevent the deformation 
of the portal of the bridge by the accumulated pressures at the 
top of the end posts, portal brackets, or portal bracing, are intro¬ 
duced. The brackets or bracing prevent any change in the 
angle between the end posts and the strut connecting their tops. 

To simplify the calculation of the stresses in the wind-brac¬ 
ing of a highway bridge, it is customary to assume that the area 
of a linear foot of the vertical projection of a truss is uniform, 
and the same for all spans up to 300 feet; for longer spans there 
is a small percentage of increase in the area per linear foot. The 
area assumed for the short spans is 10 square feet per linear foot, 
or 300 pounds pressure per linear foot, which is divided equally 
between the trusses. The same area is assumed for railroad 
bridges of less than 200 feet span. The truss connecting the 
loaded chords has in addition a moving load due to the pres¬ 
sure of the wind on a moving train or other load. This pressure 





















BRIDGES. 


467 


is assumed to be that on a surface 5 feet high or 150 pounds 
per linear foot in a highway bridge, and on a surface 10 feet high or 
300 pounds per linear foot in a railroad bridge. The effect of 
a horizontal moving load on a wind truss is similar to the effect 
of a vertical moving load on one of the main trusses. 

A bridge which has a wind truss connecting its upper chords 
is called a through bridge; the least distance between this truss 
and the floor, or the clear headroom, should be 14 feet in a high¬ 
way and 20 feet in a railway bridge. If the floor rests on the 
upper chord the bridge is a deck bridge. If the floor rests on the 
lower chord, but the trusses are not high enough to admit of 
lateral bracing between the upper chords, it is a half-through or 
pony-truss bridge. In this bridge the floor-truss transmits all the 
wind pressure; the floor-beams are usually prolonged and their 
ends are connected with the upper chords by inclined braces. 

The wind-bracing of a plate-girder or a steel-arch bridge is 
designed in a similar manner. 

The bridge is prevented from moving laterally by friction 
and by bolting it to the abutments. A strut, omitted in Fig. 
149 for clearness, connects the bases of the end posts resting on 
the same abutment and divides the wind pressure between the 
bedplates. 

The wind pressure on piers, towers, or trestles is determined 
in the same manner as the pressure on the trusses; the pres¬ 
sure per square foot is, however, assumed to be 45 to 50 pounds 
per square foot. To simplify calculation it is also assumed 
as uniform per vertical foot and to vary from 150 pounds per 
vertical foot in highway-bridge towers to 250 pounds in railway- 
bridge towers. The towers must have a factor of safety of two 
against overturning under the most unfavorable conditions; in a 
railway bridge this will be when an empty train covers the entire 

bridge. 

A high cantilever tower is rendered stable by making the 
base much wider than the top. In the Forth bridge (Fig. 140) 
the base of each tower is 120 feet wide and the top 33 feet. A 
high steel arch is made stable by making the width at the spring- 
ing-lines much wider than at the level of the crown. The Viaur 
bridge (Fig. 144) is 109 feet wide at the springing-lines and 19 
feet wide at the roadway. 


468 


CIVIL ENGINEERING . 


In a suspension bridge the wind pressures are transmitted 
to the towers by wind trusses connecting the chords of the stiff- 
ening-trusses, by the tension in the cables when these are not 
in vertical planes normal to the piers, and by auxiliary guys. 

In truss- and girder-bridges built on a curve the wind braces 
must also resist the lateral pressure due to a moving train. 

Sway-bracing.—If the vertical posts of a through bridge are 
much higher than is necessary to give the desired headroom, 
they may be connected by sway-bracing as shown 
in Fig. 150. This bracing prevents the deformation 
of the panels formed by the vertical posts, floor- 
beams, and top connecting struts, by wind pressure 
or by a load not symmetrically disposed with re¬ 
spect to the two trusses. A train on one track of 
Fig. 150. a double-track railroad bridge would be an ex¬ 
ample of such loading. Since no headroom need be provided, 
in a deck bridge each diagonal connects the top of each post 
with the base of the one opposite. 

Temperature Stresses.—Allowance is made in bridge design 
for a variation of temperature of 150 degrees. 

Bridge Erection.—The term plant is applied to the derricks and 
other machinery employed in erecting the bridge; the term jalse 
works to the trestling that supports the bridge during erection. 

Trestle-bents if not too high are framed on the ground and 
then lifted into place. High bents and towers are constructed 
in position by the use of the ordinary hoisting-apparatus. 

Plate girders and riveted Warren girders are delivered on 
the site in sections as long as can be conveniently transported 
and handled. These sections are riveted together on the site 
and the girder is then put into place by moving it on rollers 
which are supported by stringers resting on trestle-bents. A 
large girder is hauled into position by a rope attached to a cap¬ 
stan. Small girders may also be placed in position by means of 
derricks alone. A Howe truss like that shown in Fig. 142 is 
assembled on shore and placed in position in the same manner 
as a girder. 

A pin-connected truss requires the erection of a platform at 
the level of the lower chord. All the members that are attached 
to the pins of the lower chord are assembled on this platform. 





BRIDGES. 


469 


A second platform may be constructed for the upper chord or 
the pieces of this chord may be raised and united to the web 
members by means of a traveler. This is a car which runs on 
a track on the lower-chord platform and supports a longitudinal 
cantilever or crane at a level above that of the upper chord. 
The cantilever is provided with hoisting-apparatus. 

Cantilever spans are erected without false works. The canti¬ 
lever which is being constructed is supported by a similar one 
constructed at the same time on the opposite side of the tower,, 
or by a truss which has been previously constructed. The sus¬ 
pended truss connecting two cantilevers is constructed as a part 
of the cantilevers; when united at the middle point of the span 
the upper chord is severed between the cantilever and truss. 
The construction stresses in the chords of the suspended truss 
are therefore the reverse of their final stresses. This change must 
be provided for in the design. As the cantilever requires no false 
works during construction it is particularly suitable for sites 
where, on account of the depth of water or other conditions, 
false works would be very difficult to erect or maintain. 

Masonry and reinforced concrete arches are constructed on 
centers. Steel arches are usually constructed without false 
works, the character of the site being such that the semi-arches 
can be held by cables until they are united at the crown. If the 
site does not admit of this method of construction the arch is 
supported on a center. 

Suspension bridges are constructed by means of platforms, 
cars, and travelers suspended from the main and auxiliary cables. 

Drawbridges.—A drawbridge is one in which one or more 
spans can be temporarily raised or removed to allow the passage 
of vessels. They are usually constructed over navigable rivers 
with low banks. 

The draw-span may be raised vertically or moved horizon¬ 
tally, or it may be rotated about a horizontal or vertical axis. 

In the first type, or lijt-bridge, the floor is lifted vertically until 
the clearance is sufficient for passing vessels. The floor is at¬ 
tached to counterweighted cables which pass over pulleys sup¬ 
ported by steel towers. The force required to lift the bridge may 
therefore be reduced to that necessary to overcome the friction 
in the hoisting-apparatus. 


47© 


CIVIL ENGINEERING . 


In the second type the draw-span is supported by a car which 
runs on a track whose axis coincides with that of the span, or it 
is supported on a boat or ponton which is floated out of position. 

The above types are employed only where the conditions are 
extremely favorable for their construction; draw spans, as a 
rule, rotate either about horizontal or vertical axes. 

In the third type, or bascule bridge , the draw-span rotates 
about either a fixed or a movable horizontal axis and may con¬ 
sist of one or two leaves. If the span is short, hinges are fastened 
to the shore ends of the floor; these rotate about pins or bars 
attached to the abutments. The leaf is rotated by means of 
counterweighted chains or cables attached to its outer end, which 
pass over pulleys in walls or towers above the abutments. 
The drawbridge over the moat of an old fortification is of this 
type. If the span is long, the hinges are so placed that the center 
of gravity of the revolving leaf shall be at or near the hinge; 
the leaf is then a double cantilever with a short heavy shore arm 
which rotates in a well made in the abutment. The London 
tower bridge, whose draw-span is 290 feet, is of this type. Fig. 
151 represents a bascule bridge with a movable axis. The short 



arm supports a heavy counterweight W. This is the type com¬ 
monly constructed in this country and has been applied to a span 
of 275 feet in Chicago, Ill. In large-span bascule bridges the 
cables are dispensed with and the power is applied to the leaf 
at the abutments. The position of equilibrium of the leaf may 
be horizontal or inclined; the latter facilitates its rapid opening. 

In the fourth type, or wmg-bridge, the draw is in the form 
of a double cantilever with arms of equal or unequal length. 
If the span is short and a single opening is desired, the axis of 





























BRIDGES. 


47 i 


rotation may be on the abutment and the shore cantilever may 
be short and heavy. The ordinary construction, however, is 
to place the- axis on a pier, make the cantilever arms equal, and 
thus secure two equal openings. There are several bridges of 
this type in this country in which the double cantilevers are 
over 500 feet long from end to end. The longest, 520 feet, is at 
Omaha, Neb. The draw-spans of swing-bridges are supported 
by rollers which move on a circular track. 

The cantilevers may be either plate girders or trusses. If 
plate girders, they are designed first as unloaded cantilevers 
when the spans are open, and as continuous beams resting on 
three points of support when the spans are closed and loaded. 
If the bridge is a truss-bridge, the truss is usually designed as 
a double cantilever when the spans are open and unloaded, and 
as two trusses resting on two supports each when the spans are 
closed and loaded. 

Swing-bridges are usually preferred in wide rivers where 
the central pier is no serious obstruction to navigation; bascule 
bridges are preferred where the channel is narrow. 

For further information consult Waddell’s “ De Pontibus,” Foster’s 
“Wooden Trestle Bridges,” Cooper’s “Specifications for Railway and High¬ 
way Bridges,” Thacher’s “Specifications for Railway and Highway Bridges,” 
Greene’s “Arches,” Merriman and Jacoby’s “Higher Structures.” 


Illustrated Descriptions of Bridges. 


Luxemburg stone arch.Eng. News, vol. 47, p. 179 

Plauen stone arch. ‘ 5 1 , P* /4 

Topeka reinforced concrete. . “ 37 . P- io 5 an d v °l- 39 , P- 99 

Viaur steel arch. “ 44 , P- I 5 § 


Niagara steel arch. 

Forth cantilever. 

Quebec “ . 

Memphis “ . 

Poughkeepsie cantilever. 

Brooklyn suspension. 

Williamsburg “ . 

Bascule drawbridges. 

I ift drawbridge. 

Swing drawbridge. 

London Tower bridge....... 

Pecos River steel tower bridge 


“ “ “ 37 , P- 252 

Engineering, vol. 49, p. 213 
Eng. News, vol. 37, p. 252 

“ “ “ 27, p. 470 and vol. 28, p. 251 

Eng. Bldg. Record, May 1888 
Eng. News, vols. 5 and 6 
“ “ vol. 50, p. 535 

“ “ “ 45 , P- 18 

“ “ “ 3 L P- 320 

“ “ “ 3 °, P- 448 

“ “ “ 3 L P- 43 

“ “ “ 27, p. 125 


















CHAPTER XXIV. 

TRUSSED ROOFS AND FLOORS. 

Definitions. — Fig. 152. The roof-covering is the outer or 
water-proof coating of the roof. The materials employed for 



Fig. 152. 


this coating are shingles, slate, tiles, asphalt, tin, copper, lead, 
and corrugated iron. 

The sheathing is the layer of boards or other material to 
which the covering is attached. 

The rafters are the inclined beams which support the sheath¬ 
ing; they correspond to the joists in bridge floors. 

The purlins are the horizontal beams which support the 
rafters and correspond to the floor-beams of bridge floors. 

The roof-trusses are the frames which support the roof and 
transmit its weight to the walls or columns of a building. 

The wall-plates are the wooden beams which are laid on the 
top of the wall to distribute the weight transmitted by the trusses; 
if rectangular stones are used for the same purpose, they are 
called templates. 

The pitch of a roof is the angle which its plane makes with 
the horizontal; in simple roofs the pitch is also expressed as the 
ratio of the rise to half the span. The most common pitch is 
26^ degrees, or J. 


472 


























TRUSSED ROOFS AND FLOORS. 


473 


The ridge is the highest horizontal line of the roof. 

The eaves are the lowest horizontal lines of the roof. 

A shed or lean-to roof has a single plane surface ( a , Fig. 153). 




Fig. 153. 



A gable roof has two plane surfaces intersecting in the ridge 
(b, Fig. 153). 

A curb or gambrel has four plane surfaces; the two on 
each side of the ridge have different inclinations and intersect 
in a line parallel to the ridge, called the curb ( c , Fig. 153). 
The mansard is a curb roof with steep sides and a rather flat 
top. 

A hipped roof has the same inclination at the ends of the build¬ 
ing as at the sides; the inclined lines of intersection of its slopes 
are the hips (1 d , Fig. 153). 

A valley is the line of intersection of roof surfaces making 
a reentrant angle, as the line of intersection of two gables which 
are perpendicular to each other. 

Roofs may also be arched, conical, and dome-shaped. 

Construction.—The kind of roof-covering varies with the 
character of the building, the architectural effect desired, and the 
pitch of the roof. If the angle of pitch is small, the covering 
should be a continuous sheet, as asphalt, tin, etc.; otherwise 
the covering may be composed of overlapping plates, as slate, 
tiles, etc. 

The sheathing is usually inch boards, covered by layers of 
tarred paper or similar materials; when a fire-proof construction 
is desired, the sheathing consists of slabs of terra cotta or other 
fire-proof material supported by a metal framework. 

Wooden rafters are beams 2 to 3 inches thick and 4 to 10 
inches deep. They are spaced from 16 to 24 inches, center to 
center, for ordinary loads; these distances may be increased or 
diminished. If the wooden rafters rest directly upon the wall- 
plates and not on purlins, the opposite rafters are connected 


















474 


CIWL ENGINEERING. 


about midway between the wall-plates and ridge by a horizontal 
brace called a collar-beam. In metal roofs the rafters are I beams 
or other structural steel forms. The I beams are spaced at 
wider intervals than wooden rafters, and in fire-proof construc¬ 
tion are connected by inverted tees which support the fire-proof 
sheathing. 

The purlins are usually spaced about 8 to io feet, as wider 
spacing requires the rafters and purlins to be inconveniently 
large. They are either square wooden beams or one of the 
forms of structural steel beams. The rafters are sometimes 
omitted and the sheathing fastened to the purlins themselves; 
the purlins must then be spaced as rafters. 

The trusses are made of wood, of wooden compression and 
steel tensile members, or wdiolly of steel. The compression 





chord corresponds in general outline to the roof itself; the ten¬ 
sion chord may be a right or broken line. The web members 
may all be inclined, as in a Warren truss; the compression 
members may be vertical and the tension members inclined, as 
in the Pratt truss; the compression members may be inclined 
and the tension members vertical, as in the Howe truss; or the 
upper chord may be trussed, thus forming a Fink truss (Fig. 
154). * The panel points of the upper chord are usually at the 
purlins. 

In wooden trusses the members are all of the same thick¬ 
ness and differ only in depth; they are united by mortise and 
tenon or notched joints, strengthened by straps or bolts. Steel 
trusses may have pin-connected joints along the lower chord 
or they may be riveted throughout; the latter is the more com¬ 
mon construction. In the riveted truss all the members are usu- 















TRUSSED ROOFS AND FLOORS. 


475 


ally angles in pairs riveted to connecting plates and fastened 
at intervals to each other. 

Like the bridge-truss, one end of the roof-truss is anchored 
firmly to its support; the other end rests on a smooth plate or on 
rollers which permit longitudinal motion but do not permit lateral 
or vertical motion. 

Arched roofs may have their trusses in the form of the bow¬ 
string truss, or they are in the form of an arch with parallel chords, 
and are usually hinged at the springing-lines and crown. 

Loads.—The dead load of a roof is the weight of the material 
of which it is made and in addition any permanent weights sus¬ 
pended from its trusses. The live load is the pressure of the 
wind, the weight of the snow which may lie on it, and any tem¬ 
porary weights which it may have to support. The dead load 
is approximately as follows for roofs whose span does not exceed 
75 feet. The weight is that of the actual roof area. 

Shingles, tin, or equivalent. 10 lbs. per square foot. 

Slate on boards. 15 “ “ “ “ 

Fire-proof construction of steel and 

terra-cotta tile. 40 “ “ “ “ 


The live load , excluding the wind, is usually assumed as 25 
pounds per square foot of roof surface. Flat roofs used as 
floors are designed for floor loads. 

In computing the wind pressure on a roof the action line of 
the wind is assumed to be horizontal and its intensity is gener¬ 
ally taken as 30 pounds per square foot. 

Let F = total pressure of the wind on a vertical plane nor¬ 
mal to its action line; 

N = normal pressure on the same plane when inclined 
to the horizontal at an angle <f>\ 

H = the horizontal component of N, or N sin 0; 

V = vertical component of N , or N cos <j>. 

The relation of F and H was determined by experiment to be 


*H = F sin f- 842 COS ^ or 


f H = F 


2 sin 2 (f> 
1 -I-sin 2 <j> 


* Hutton. 


f Duchemin. 








476 


CIVIL ENGINEERING. 


The first gives the ratios of N , H , and V to F for different 
values of </> as follows: 


<f) in degrees. 

IO 

20 

30 

40 

50 

60 

70 

80 

90 

N 

F 

.24 

•45 

.66 

.83 

■95 

1.00 

1.02 

1.01 

1.00 

H 

F 

.04 

•15 

•33 

•53 

•73 

.85 

.96 

•99 

1.00 

V 

F 

.24 

.42 

•57 

.64 

.61 

•50 

•35 

• i 7 

0.00 


The normal pressure N must be resisted by the roof con¬ 
struction, which transmits its effects to the walls. 

The horizontal pressure H tends to move the roof horizon¬ 
tally and must be resisted by its anchorages. It also tends to 
overthrow the walls which support the roof. 

The vertical component V must be added to the dead and 
other live loads in computing the strength of the walls. 

Merriman gives the following formula for determining the 
ap proximate weight of a roof-truss. It is employed in computing 
approximate dead load. 

W = \al + T V for a wooden truss; 

W = \al -f T V for a steel truss; 

in which W = weight of one truss in pounds; 
a = truss interval in feet; 
l = span of truss in feet. 

Floors. 

The floor of a building is constructed in the same general 
manner as the floor of a bridge and consists of flooring, joists, and 
girders. Its essential qualities are strength and stiffness. 

According to the method of construction floors may be 
divided into wooden, I-beam, and reinforced concrete floors. 

Wooden Floor.—In the wooden floor the flooring is made of 
narrow-tongued and grooved boards which are supported by 
wooden joists. 











































TRUSSED ROOFS AND FLOORS. 


477 


To secure stiffness as well as strength, the width of the joists 
is small as compared with their depth; the width varies from 
2 to 3 inches and the depth from 6 to 14 inches. They are spaced 
from 12 to 16 inches. To hold the joists in position they are 
bridged at the middle point of short spans and at intervals of 8 
feet in long ones. The bridging consists of vertical cross-bracing 
composed of 1 by 3 inch pieces, which is inserted between con¬ 
secutive joists. The joists are supported by the walls of the 
building or by girders. If the upper surfaces of the joists and 
girder are to lie in the same horizontal plane, the joists are 
notched into the girders or are supported by iron stirrups which 
are fastened to the girders; the latter method is the better, as 
the total cross-section of both joists and girder are preserved. 

The girders are rectangular beams which rest on the walls 
of the building or upon columns or pillars. If the span is great, 
wooden girders are replaced by steel I beams. 

I-beam Floor.—In this floor the joists are steel I beams, and 
the girders steel I beams, plate or box girders. The joists and 
girders are fastened together by angles riveted to the webs of 
the girders. The joists are connected by arches or plates of 
incombustible material such as bricks, tile blocks, or concrete. 
Above the arches or plates the floor is leveled with cinder con¬ 
crete and in this concrete bed are placed wooden strips to which 



Fig. 156. 


the flooring is nailed. The flooring may also be made of tiles 
or other fire-proof material. The fire-proof floor differs from 
the semi-fire-proof in having its joists and girders protected from 
the flames of a fire underneath it, by a ceiling or other shield 
















47 8 


CIVIL ENGINEERING. 


of fire-proof material. (Figs. 155 and 156.) The construc¬ 
tion of the brick and tile-block floors is shown in Figs. 155 and 
156; the depth of the brick arch for spans of 5 feet is about 4 
inches; the flat tile-block arch is deeper. The thrust of the 
arch is neutralized by ties connecting the consecutive joists at 
intervals of 8 feet. When the blocks are laid as in Fig. 156 the 

method is called side construction ; when laid 
as in Fig. 157 it is called end construction. 
The side construction is usually preferred, 
although the block is stronger longitudi¬ 
nally than laterally, because the arch is 
more easily constructed. Holes may be 
F IG . 157. made through these arches for the passage 
of flues and pipes. 

The joists may also be connected by arches of concrete or 
plates of reinforced concrete (Figs. 158 and 159). The arches 




Fig. i 58. 



Fig. 159. 

and plates are supported during construction by platforms sus¬ 
pended from the steel-floor joists. All these forms of floors will 
safely support the loads given below. The fire tests required 
of fire-proof floors are given in standard building regulations. 

Reinforced-concrete Floors. — In fire-proof buildings the 
I-beam floor is now being replaced by the reinforced-concrete 
floor because of its cheapness and fire-resisting qualities. It 
differs from the I-beam floor in having no rigid framework of 
steel as its basis; only a minimum amount of metal is employed 
in reinforcing the floor-beams, joists, etc., usually not exceeding 

































TRUSSED ROOFS AND FLOORS. 


479 


one per cent of the cross-section. The floor proper is a large 
slab of reinforced concrete which is supported on the joists and 
covered by any desired floor surfacing. By means of the rein¬ 
forcing bars or wires and the concrete, all the parts of the floor 
system are firmly attached to the supporting columns and to each 
other and made into a monolithic mass. Its cheapness results 
from the small amount of reinforcement required, and its fire- 
resisting qualities from the thorough coating of the metal by 
concrete. During construction a reinforced-concrete floor must 
be supported throughout by a platform. Many different designs 
of these floors have been patented and are being constructed. 

Loads.—Each member of the floor system must be designed 
to resist a dead load equal to the weight of the member and the 
weight of so much of the floor as is transmitted to it by the other 
members, and a live load which depends on its probable loading. 
The live load is assumed to be uniformly distributed. 

Dead Load.—The weight of a wooden floor including its 
joists is about io pounds per square foot; with the attached 
ceiling 20 pounds. 

The weight of a brick or block floor arch in pounds is about 
four times the depth of the arch in inches, and may vary from 
25 to 45 pounds per square foot. If the weight of the I beams, 
flooring, and ceiling are added, the weight will be increased to a 
total of 75 to 100 pounds. The floors of the government printing- 
office in Washington, which were designed for live loads of 300 
pounds per square foot, were computed for a dead load of 125 
pounds per square foot. In this building the supporting columns 
are in rows 35 feet centers, and the columns in each row are at 
intervals of 12 feet. 

Live Load.—According to standard building regulations the 
live loads for which floors must be computed are: 


Dwellings. 

... 40 to 

60 

Schools. 


100 

Stables. 

... 75 “ 

100 

Public assembly halls. 

... 90“ 

120 

Warehouses. 

... 120 “ 

25° 


< ( 
i ( 


i t 
< i 
t 1 


< t 
t € 
t < 
t ( 


i i 
( 6 
f i 


Some of these loads are considered by competent engineers 
as not complying with the conditions of practice and the follow¬ 
ing table (page 480) has been rec ommended to replace it * 

* Schneider in Proceedings of Am. Soc. of Civil Eng’rs, Sept. 1904. 










480 


CIVIL ENGINEERING. 



Live Loads, in Pounds. 

Classes of Buildings. 

Distributed 

Load. 

1 

Concentrated 

Load. 

2 

Load per 
Linear Foot 
of Girder. 

3 

Dwellings, hotels, and apartment- 
houses. 

40 

2,000 

500 

Office buildings. 

40 

5,000 

1,000 

Assembly rooms with fixed seats, like 
theaters, churches, schools, etc.. . 

40 

5,000 

1,000 

Assembly rooms without fixed seats, 
like ballrooms, gymnasia, armo¬ 
ries, etc. 

80 

5,000 

1,000 

Stables and carriage-houses. 

70 

5,000 

1,000 

Ordinary stores and light manufac¬ 
turing. 

40 

8,000 

1,000 

Sidewalks in front of buildings. 

Warehouses and factories. 

IOO 

from 120 up 

10,000 

Special 
( ( 

Special 

i i 

Charging floors for foundries. 

“ 300 ‘ ‘ 

Power-houses, for uncovered floors. . 

“ 200 “ 

The actual weights of 



engines, boilers, stacks, 
etc., shall be used, but in 
no case less than 200 lbs. 
per square foot. 


The live load to be employed is the one in columns 1, 2, or 
3, which produces the greatest stress in the member considered; 
the distributed load is the weight of people, furniture, etc.; the 
concentrated load is that of a safe, which may be omitted in 
computing the dimensions of floor members of simple dwellings. 
In estimating the live load transmitted from the floors to the 
columns of high buildings a reduction of 5 per cent is made for 
each floor below the top one until this reduction reaches 50 per 
cent. This corresponds to the reduction made in the live load 
for long-span bridges. 

For further information consult Fowler’s “Specifications for Steel Roofs 
and Buildings”; Schneider’s “Structural Design of Buildings”; Transactions 
American Soc. of Civil Engineers, 1905; Kidder’s “Building Construction,” 
Part I; Merriman and Jacoby’s “Higher Structures.” 



















CHAPTER XXV. 

HIGHWAYS. 

Definitions. A highway is a line of communication over 
which the public has a right of way. The ordinary highway 
is designed for the use of wheeled vehicles. 

A street is a highway of a village or city designed for general 
use; an alley is a narrow highway designed for the use of the 
owners of abutting property. 

A road is a country highway which connects villages and 
towns. 

The carriageway is that part of a highway designed for the 
use of vehicles. 

The sidewalks are the parts of the highway reserved for 
pedestrians. 

A pavement is a layer of hard material placed on the natural 
surface of a carriageway or sidewalk. The term is usually con¬ 
fined to block or sheet coverings. 

A macadam or broken-stone carriageway is one which is covered 
by a layer of stones broken into small fragments. 

Gutters are shallow depressions along the sides of the carriage¬ 
way of a street. They are usually paved to catch the surface 
drainage of the street and conduct it to the sewers. 

A side ditch is a deep depression along the side of the car¬ 
riageway of a road. It catches the drainage of the road and 
of the adjoining lands which slope towards the road and con¬ 
ducts it to the nearest natural watercourse intersecting the road. 

A curb is a line of narrow stones which supports one side of 
the pavement or gutter of a street and separates the carriage¬ 
way and sidewalk. 

The axis of a highway is the line of its surface which bisects 

481 


482 


CIVIL ENGINEERING. 


it longitudinally. The axis of the carriageway is a similar line 
which bisects the carriageway. 

The trace of a highway is the projection of its axis on a hori¬ 
zontal plane. 

The profile is the development of the axis and trace on a 
vertical plane. 

The grade of any part of the axis of the carriageway is its 
inclination to its trace. It is usually indicated by its rise per 
mile or per hundred feet, assuming the grade to be uniform for 
this distance. A grade of t To is called a 1 per cent grade; a 
grade of T f q a 2 per cent grade, etc. 

A transverse or cross section of a highway is the intersection 
of its surface by a plane normal to its axis. 

The crowning of the carriageway is the vertical height of 
its axis above its sides. 

Profile. —The profile of a highway is the natural profile of 
the country over which the road passes modified by cuts and 
fils. The cuts and fills together reduce both the irregularities 
in the profile and the inclination of its slopes or grades. The 
maximum desirable grade on a highway is such that a horse can 
pull his ordinary load up the grade without loss of speed and 
without undue fatigue. This grade will depend therefore on the 
tractive power of the horse and on the load and resistance of 
the vehicle he draws. 

Tractive Power of a Horse. —The tractive power exerted by 
the horse at any moment is approximately measured by the 
tensile stress in the traces. The working tractive power is that 
which he can exert throughout a working day when he moves 
at a walk on a level road. This is now generally assumed as 
120 pounds for a horse weighing 1000 pounds, or is about one- 
eighth his weight. * His maximum tractive power is that which 
he can exert for a few moments, as in starting a load; this is at 
least twice his working tractive power and is equal to one-fourth 
to one-third his weight. 

The amount of useful work which can be performed by a 
horse is the product of the tractive power by the distance over 
which he can exert this power daily. It is generally assumed 
that a horse can, without undue fatigue, walk 20 miles a day on 
a good level road, and at the same time pull a wagon whose con- 


HIGH IVAYS. 


483 


stant resistance is 120 pounds. This is equivalent to 12,672,000 
foot-pounds of work. The normal horse-power employed in 
mechanics is based on a horse whose tractive power is 150 
pounds moving at the rate of 2J miles per hour or 220 feet per 
minute; this gives 33,000 foot-pounds per minute or 15,840,000 
per day of eight hours. 

If required to exert a greater tractive force than 120 pounds, 
the horse must have occasional rests, which will decrease his 
daily rate, the distance he travels, and his daily useful work. 
The limit is reached when his load is so heavy that he can no 
longer start it. 

If the horse is required to move more rapidly than a walk, 
his tractive power and his daily useful work are both diminished. 
The limit is reached in the race-horse who pulls a vehicle whose 
resistance is made as small as possible. 

The maximum tractive power of a horse on a slope is less 
than on a level; the action line of his weight is inclined to the 
surface, which reduces the friction of his shoes and also reduces 
his pulling moment. The reduction in his tractive power in¬ 
creases with the smoothness and hardness of the surface. Since 
part of his muscular effort is expended in raising his own weight, 
his daily useful work will be less and his non-available work will 
be greater than on a level road. 

Weight of Load.—In this country the vehicle in common 
use is the two-horse farm wagon whose weight is about 1200 
pounds and which carries a load of 2000 pounds. The weight 
pulled by each horse is therefore about 1600 pounds. Our 
army wagon is a four-horse or six-mule wagon whose weight is 
2000 pounds and its maximum load 4000 pounds; the total 
weight is therefore 1500 pounds per horse. It will be observed 
that the weight of the vehicle in each case is about one-third the 
total weight. On city pavements the loads are usually much 
heavier. 

Tractive Resistance of Vehicles.—The tractive resistance of 
a vehicle on a level surface is measured by the force which must 
be applied to the vehicle, parallel to the surface upon which it 
rests, to move it at a uniform rate of speed. This resistance is 
the sum of the frictional resistance between the axles and hubs, 
and the resistance at the surface of the wheels caused by irregu- 


484 


CIVIL ENGINEERING. 


larities in the surface of the road. For each wheel the total re¬ 
sistance may be expressed by the formula 


R = jW 


aW 
v 7 r 5 


in which R = total resistance; 

/ = coefficient of friction of surfaces of axle and hub; 

IF —weight on wheel; 
r =radius of wheel; 

<z=a coefficient dependent on character of road sur¬ 
face. 

aW 

The term jW is the resistance at the axle, and is the 

\ r 

resistance due to irregularities in the surface of the road 
determined by experiment. 

If a dynamometer be placed between the traces and the 
vehicle, and the vehicle be drawn at different rates of speed - 
over level roads with different surfaces, the total resistance of 
the vehicle will be in the form 


R' = I'W', 

V 

in which R' = total resistance in pounds; 

W' = total weight of vehicle and load in pounds; 

/' = coefficient dependent on the character of the road 
surface and speed. 

Many experiments have been made to determine the value 
of /' corresponding to the different classes of roads and pave¬ 
ments. While there is a general agreement in the results ob¬ 
tained by the different investigators, there is also a consider¬ 
able variation, due to the differences in the vehicles used, to 
the difficulty of driving horses at a uniform rate, and to the fluc¬ 
tuations in the dynamometer readings caused by slight irregulari¬ 
ties in the surface. The values given below can therefore be 
considered approximate only, and for roads in fair condition when 
the vehicle moves at the rate of 2J miles per hour. 




HIGH WA YS. 


485 


l 


Surface Pounds per Ton Tractive Resistance 

of 2000 lbs. Value of 

Steel railway. 8 I / 2 5° 

Steel tramway. 15 1/133 

Hard sheet asphalt. 25 1/80 

Soft sheet asphalt. 70 1/30 

Brick. 25 1/80 

Stone blocks.’. 40 1/50 

Macadam. 60 1/33 

Gravel road. 90 1/22 

Earth road. 125 1/16 

Loose gravel. 300 1/7 

Loose sand. 400 1/5 


Assuming the working tractive power of the horse at 120 
pounds, if he can secure a good foothold, he can, on a level, pull 
with equal ease 

15 tons on a railway. 1^ tons on a gravel road. 

8 tons on a tramway. 1 ton on an earth road. 

4! tons on hard sheet asphalt or brick. f ton on loose gravel. 

3 tons on a stone-block pavement. ^ ton on loose sand. 

2 tons on a macadam road. 

If the surface of a carriageway is smooth, the resistance is 
independent of the speed; if it is rough, the resistance increases 
with the speed. 

In pulling a vehicle up a grade, the tractive force applied must 
overcome the same resistances as on a level, and in addition must 
also overcome the component of the weight of the vehicle parallel 
to the slope. The total tractive force must therefore be 

T = j'W cos cf> + W sin <j ), 


in which <f> is the angle of inclination of the slope. Since this 
angle is never very large, no great error is introduced if we assume 

1 h 

cos (f> equal to unity, and sin cj) equal to tan <£ or equal to y, 


in which h is the rise and / is the horizontal length of the slope. 
Making these substitutions we have 


Wh 

T = j'W +—y. 


The value of —7-, the tractive resistance due to the slope 

i 

alone, is the same for all roads and varies directly with the 

h 

weight of the vehicle and the inclination -y. We may therefore 


form the following table: 














486 


CIVIL ENGINEERING. 


Resistance per Ton (2000 lbs.) of Load due to Grade. 


1% grade. 

... 20 lbs. 

2% “ . 

... 40 “ 

3 % “ . 

... 60 “ 

4 % “ . 

... 80 “ 

5 % “ . 

... 100 “ 


6% grade.. . . 

. . . 120 lbs. 

7 % “ .... 

. . . 140 ‘ ‘ 

8% “ .... 

... 160 “ 

9 % “ .... 

... 180 “ 

10% “ .... 

... 200 * * 


These resistances must be added to those given in the pre¬ 
vious table to determine the total tractive resistance of a vehicle 
which is being pulled up a slope. It will be seen that the relative 
effect of a slope is greater on a good road than on a poor one. 
Thus on a hard asphalt pavement the total resistance on a 1 per 
cent grade is nearly double the resistance on the level, while 
on an earth road the total resistance on a 1 per cent grade is only 
one-sixth greater than on the level. The better the road surface, 
therefore, the more important it is to reduce the grades. 

Wh 

In moving down the slope the force — j~ acts in the same 


direction as T, and the expression for T becomes 


T = j'W 


Wh 

l * 


If j'W = 


Wh 
l ’ 


then T reduces to zero. 


Therefore in moving 


down a slope whose inclination, —, is equal to the coefficient /', 

the horse will not be required to exert any tractive force. In mov¬ 
ing up the same slope his tractive force must be 2 j'W, or double 
the amount necessary to move the load on a level surface. 

If the road has a macadam surface, for which / is equal to A-, 
no tractive force is required to pull a load down a slope of ^ 
or jfg-, and a tractive force twice that required to pull a load 
on the level will be required to pull the load up this slope. For 
these reasons a slope of called a 3 per cent grade, is consid¬ 
ered the ideal maximum grade of a macadam road. The angle /' 
is called the angle oj repose, and, as will be seen from the table, 
its value depends on the character of the surface. 

Maximum and Minimum Grades.—In road construction it 
is necessary to fix the maximum and minimum grades in the 
profile. On account of the expense of construction it is impossible 
to adopt the angle of repose as the maximum grade on all roads. 
In France, where the roads are all macadamized or paved, the 















HIGH IVAYS. 


487 


macadamized road is the standard, and its angle of repose, 3 per 
cent, is adopted for all roads of the first class. For roads of the 
second class the maximum grade is 4 per cent, and for those of 
the third class 5 per cent. The allowable loads on first-class 
roads are about twice those adopted in this country. 

In this country a road may be classed as a first-class road when 
its maximum grade does not exceed 5 per cent; a second-class 
road when this grade does not exceed 7 per cent; and a third- 
class road when its maximum grade does not exceed 10 per cent. 
A grade of 5 per cent corresponds to the angle of repose of a 
smooth earth surface. 

The minimum grade in the profile is that which will permit 
of satisfactory drainage. This is usually fixed at about 0.8 to 
1 per cent or T ^j to t -J-q. 

Cross-section.—In discussing the cross-section of a road a 
distinction must be made between the country roads, where there 
are few pedestrians, and the village and city streets, where pro¬ 
vision must be made for them. 

Roads.—The total width of a road is the width of the right 
of way, or the strip of land over which the public has the right 
to pass. This width is fixed by law, and in this country varies 
between 2 and rods, but is commonly 3 or 4 rods, or 49J to 66 
feet. The width of the right of way of every important road 
should be sufficient to permit of the construction of a carriage¬ 
way 26 to 30 feet wide in cuts and on embankments. On un¬ 
important roads its width should be sufficient- to allow of the 
construction of a carriageway 20 feet wide. A width of 3 rods 
should therefore be the minimum for the former, and 2 rods for 
the latter. An unimproved road must be made wider than a 
macadamized road, since in wet weather the former is soon cut 
up and becomes impassable if the travel is confined to a narrow 
track. An unimproved road should also be freely exposed to 
the heat of the sun and should not be shaded by overhanging 

trees. 

In France the width of the carriageway and not the right of 
way is fixed by law; an order of the royal council in 1776 divided 
the public highways into four classes and fixed the width of road¬ 
way of each. 

The first class comprised the great roads which traversed 



4 88 


CIVIL ENGINEERING. 


the kingdom and connected the capital with the principal cities, 
ports, and industrial centers. These are now the national high¬ 
ways (routes nationales). 

The second, the roads which connect the principal provinces, 
cities, ports, etc., with each other. These are now the depart¬ 
mental roads (routes departmentales). The third and fourth classes 
were the roads which connect the towns of one province with 
those of an adjacent one, and also the roads which connected 
the towns and villages of each province. These are now the local 
roads (chemins vicinaux and are divided into three classes. 

Exclusive of the width of ditches and side slopes, the width 
of first-class roads was fixed at 45 feet, the second at 38 feet, 
the third at 32 feet, and the fourth at 25 feet. 

The general form of the cross-section of these French roads 
is shown in Fig. 160. 



Fig. 160. 


The normal width of the different parts, as actually con¬ 
structed, is given in the following table: 



Width in Feet. 

Total. 

Macadam. 

Each Unpaved 
Strip. 

National highways. 

16.4 to 22.9 
13.1 to 16.4 
9.8 to 16.4 

8.2 to I I.5 

4.9 to 8.2 
4.9 to 6.5 

32.8 to 45.9 
26.2 to 32.8 
19.6 to 26.2 

Departmental highways. 

Local roads. 



The paved portion of the carriageway is designed for the 
use of heavy vehicles at all seasons of the year, and for light vehi¬ 
cles when the unpaved portions are in bad condition. The least 
width of the paved strip is 10 feet, which is sufficient for a single 
vehicle. A width of 16 feet is sufficient for two vehicles. Since 
the introduction of railroads the national highways have lost 
their importance, and a paved strip 16 feet wide is considered 
sufficient for any road. 

The unpaved strips are used by light vehicles, serve as stor¬ 
ing places for road materials, and if the road is of sufficient width 


















HIGHWAYS. 489 

are planted with rows of trees. The trees should leave an unob¬ 
structed carriageway at least 30 feet wide. 

The curve of cross-section is either a circle or ellipse whose 
height at the center is about ^ of the width of the carriageway. 
If the road is always kept in thorough repair, the crowning is 
reduced. 

The side ditches have sufficient capacity to carry off the rain¬ 
water which falls on the road and that which flows from the 
adjacent lands toward it. If the road is raised above the sur¬ 
face of the country, the ditches may be omitted. If the road 
follows the surface of a slope and is partly in fill and partly in 
cut, only the ditch on the side of the cut is retained (Fig. 161). 



The cross-section of a side-hill road may also be made a plane 
surface sloping towards the center of curvature; this prevents 
erosion of the exterior slope by the drainage from the road, and 
also lessens the danger of accidents when vehicles pass over the 
road at high speed. 

The sides of cuts and fills are given the natural slope of the 
earth of which they are formed, or a less inclination. They are 
sodded or planted with some protecting growth. The highways 
of other European countries differ in no essential features from 
those of France. 

Streets.—The cross-section of a street differs from that of 
a road in having sidewalks; this necessitates the replacing of the 
ditches by shallow gutters. When a street is macadamized or 
paved the pavement is not ordinarily confined to a strip in the 
center, but extends to the gutter or curb on each side. The 
form of cross-section of a street is shown in Fig. 162.* 


* This is a section of a street in Washington, D. C. 









49 o 


CIVIL ENGINEERING. 


The width of the carriageway is about one-half that of the 
right of way on streets 60 feet or less in width; the ratio then 
increases and is about three-fifths on a street ioo feet in width. 
These ratios may be varied also with the importance of the street 


Level Line 

< - - 35 0' - ^ 


O 

oi 

Slope /ts 

-5.05 7 --5.05^- ^\< -5.05-- 

$3 3 S g g 

n »-> r - 1 *j 

Vitrified J Asphalt v ▼ 

ban'll— 

- Concrete - 

Gravel Sgggg 



Fig. 162. 


as a thoroughfare. A carriageway over 30 feet wide is uselessly 
expensive on a street which is subjected to moderate travel. 

Location and Construction of Roads. 

General Considerations.—To determine the best position of 
a road which is to connect any two points, it is essential to study 
the topography of the country between them either with or 
without the aid of topographical maps. 

If no maps are available, the routes which seem to be the 
most favorable from observation or inquiry must be carefully 
reconnoitered. This may be done with a sketching-case, watch, 
and aneroid barometer. The selection of the routes to be ex¬ 
amined are governed by the following considerations: 

1. The streams of a country are its lines of most uniform 
declivity. The valleys formed by its streams are therefore the 
natural locations for the network of roads by which any country 
is traversed. 

2. A wide valley is a better location for a road than a narrow 
one, since the road can be constructed at a greater distance from 
the banks of the stream which drains it. The road therefore 
requires less expensive bridges over the tributaries and is less 
liable to be injured in time of floods. 

3. The divide between valleys is usually lowest where the 
source of a tributary of the stream draining the one is nearest 
the source of a tributary of the stream draining the other. These 
gaps or passes are the natural locations for the roads connecting 
the network of roads of one river valley with those of an adjacent 
one. 

































HIGH IVA YS . 


49 1 


4. In a mountainous country the ridges separating river 
valleys are narrow and rugged; the roads are therefore confined 
almost wholly to the valleys. 

5. In a rolling country the ridges are rounded and regular. 
The roads may therefore follow both the ridges and the valleys. 
The ridge roads have as a rule steeper grades, but have fewer 
bridges and culverts and are more easily drained. 

6. In a flat country the roads are generally straight, since 
the expense of construction is dependent principally on the 
length of the road. 

Trial Lines.—In reconnoitering the country to locate a new 
road it is well to begin first at the highest point, so as to have as 
extended a view as possible of the country to be traversed. In 
moving down slopes the angle of declivity of the road should be 
tested with a clinometer wherever the natural slope of the ground 
seems to be equal to or greater than the maximum desirable 
slope of the finished road. The elevations of the salient and 
re-entrant angles of the profile may be determined by the aneroid 
barometer, and the distances by pacing, by odometer readings, or 
by any other convenient method. 

From the reconnaissance maps and the plotted profile it is 
possible to determine the best of a number of trial lines which 
have been examined. 

If a contoured map of the country is available, one or more 
trial lines may be drawn on the map, and the profiles of these 
lines constructed for further study. The vertical distance 
between the contours being known, it is easy to draw on the 
map lines which have a less inclination than the maximum 
adopted grade. Thus if the vertical distance between contours 
is 10 feet, and the maximum grade is 5 percent or fa, any straight 
line whose intercept between consecutive contours measured by 
the scale of the map is more than ten times twenty, or 200 feet, 
will represent a line of the surface of the ground whose inclina¬ 
tion to the horizontal plane is less than 5 per cent, or fa. The in¬ 
clination of the proposed road at any point may be determined 
by dividing the length of its trace between consecutive contours 
by the vertical distance between contours. 

Marking Adopted Line.—Having selected the best of the trial 
lines, the next step is to stake it out on the ground. With a 


49 2 


CIVIL ENGINEERING. 


transit, and chain or stadia, the general line of the axis of the 
road is laid out upon the ground as a broken line, and a stake 
is driven at every ioo feet and at every angle. The elevations 
of the ground at these stakes and at the salient and re-entrant 
points in the profile, above an assumed datum plane, are de¬ 
termined with a level. At every stake, and at intermediate points 
if the surface is very irregular, transverse profiles are taken of 
the entire right of way. 

Plotting. —The approximate axis of the road is now carefully 
plotted to a convenient scale, and the straight lines or tangents 
are connected by curves, usually circular. 

The rules governing curves on the French highways are: 

1. Make the radius of curvature at least ioo feet if possible. 

2. Avoid sharp curves on steep grades. 

3. Separate curves of opposite curvature by a straight track 
at least 60 feet long. 

On mountain roads, where sharp curves are unavoidable, 
the road is widened at the curve as shown in Fig. 163, and the 
curve itself is made a horizontal platform. 



The profile is plotted on profile or cross-section paper to 
convenient scales. In order to magnify the irregularities the 
vertical scale, or scale of heights, is about ten times the horizon¬ 
tal scale, or scale of distances. The transverse profiles are plotted 
either above or below their actual positions on the profile as 
shown in Fig. 164. From a study of the plan and profiles it is 
possible to determine what slight changes, if any, should be made 
in the position of the axis of the road in plan. If no change is 
made, the axis of the road is fixed in the vertical plane by drawing 
on the profile a line, usually in red, which shall make the sum of 



HIGH IVAYS. 


493 


the areas above it which must be excavated, equal to the sum 
of the areas below it which must be filled, to bring the road to 



Fig. 164. 


proper grade. In the profile of this axis the straight lines are 
connected by curves. This balancing of cuts and fills for the 
entire road and for short lengths reduces the cost of construction. 

In making this change care must be taken not to increase 
the grade at any point above the maximum adopted. 

Cuts and Fills. —To determine the volume of each cut and fill, 
each transverse profile is carefully plotted to scale on cross-section 
paper, and on it are drawn the surface and side slopes of the 
road. In Fig. 165 the assumed width of the right of way is 60 























CF 

toss- 

5ECTI 

DN Of 

; ST A 

KE 6 











' 



30 ft. 


- ^ 










/ 



irea 137 sq. ft 


D 

s+4.0 



+ 2.0 





B 


-2.0 

"b 

J 

/ 



A 

>.0 





































Fig. 165. 


feet; let it be required to determine the area of the fill at stake 6 
if the road is graded to a width of 30 feet. Through the point A 
of Fig. 165, which is a point of the profile, draw the line BB, or 
the line of transverse slope. At A erect a perpendicular and lay 











































































































494 


CIVIL ENGINEERING. 


oft on it 4 feet, the elevation of the finished grade above the 
surface of the ground. Draw the horizontal line CD, making it 
30 feet in length, and draw the side slopes through C and D . 
The area thus inclosed will be the area of the fill. In this diagram 
the same scale is employed for vertical and horizontal distances. 
The area of the fill may be computed in several ways. It may 
be done by dividing the total area into a number of triangles or 
other figures whose areas are readily computed; it may be done 
by dividing the surface into a large number of small trapezoids 
by equidistant parallel vertical lines. If the distance between the 
lines is represented by d, and the length of the lines by l, V, l", 
etc., the area will be d(l + l' + /" + etc.). The total length / + /' + /", 
etc., may be measured with a map measure. The accuracy of 
the result will increase as d decreases. The area may be ascer¬ 
tained by counting the number of squares which it covers on the 
cross-section paper and then multiplying this number by the 
area of a square. The area of a profile may be determined by 
means of a planimeter. 

The volume of the cut or fill between any two consecutive 
cross-sections is obtained in one of three ways: by the use of 
the prismoidal formula in which the sum of the areas of the end 
sections plus four times the mid-section is multiplied by one-sixth 
of the distance between end sections; by multiplying the half¬ 
sum of the end areas by the distance between them; and by 
multiplying the s ction midway between the end sections by the 
distance between the end sections. The first gives a correct 
result if the volume is regular, the second an approximate result 
larger than the first, and the third an approximate result smaller 
than the first. If by the three method we compute the volume 
of a frustum of a pyramid with square bases 2X2 and 4X4 feet, 
and a height of 10 feet, the volumes given by the three methods 
will be 93 J, 100, and 90 cubic feet, respectively. The second 
method is used because of its simplicity, unless the work is to 
be done by contract and the ayment is to be based on the com¬ 
putation; then the first method is employed. 

Knowing the volume of each cut and fill, the number of 
square yards may be written on the profile below each cut and 
fill, and from the resulting diagram the most economical direc¬ 
tion of haul of the material can be determined. A fill which is 


HIGHWAYS. 


495 

too far from the adjacent cuts to be economically made from 
these cuts is made from a borrow-pit which is secured in its vicinity; 
a cut which cannot be economically placed in one of the adjacent 
cuts is deposited in a spoil-bank on ground secured in its vicinity. 

Earth which is excavated from a cut will shrink in volume 
about io per cent when placed in an embankment which is thor¬ 
oughly compressed. Rock which is excavated from a solid ledge 
will expand in volume about 50 per cent when placed in an em¬ 
bankment. These volumes of the earth-fills must therefore be 
increased 10 per cent, and those of rock must be decreased 50 
per cent, in the diagram above described. In contract work the 
amount of compression or expansion which is to be considered 
in computing volumes should be stated. 

If the cuts and fills thus computed balance as closely as seems 
desirable, the location of the road in plan and profile may be 
considered fixed, and it only remains to complete the work on 
the site by staking out the curves, marking the height of the fill 
or the depth of the cut on each axial stake, and marking the limits 
of the excavations. 

The curves are usually put in by eye; where accuracy is 
desired a curve of small radius may be described from its center 
by using the chain as a radius, and one of large radius by angles 
of deflection as described in manuals for railroad work. The 
height of fill and depth of cut, as well as the horizontal distance 
from the axial stake to the limit of the excavation, are taken from 
the transverse profiles or cross-sections. 

Execution of the Earthwork. 

Having cleared the right of way of trees and brush, which 
should not be thrown into the depressions as part of the fills, the 
material of the cuts is loosened by plows, excavated by hand or 
by machinery, and moved to the fills in barrows, drag-scrapers, 
carts, wagons, wheeled scrapers, or dump-cars. Steam exca¬ 
vators are employed only when the cut is very deep. Barrows 
and drag-scrapers are employed when the material is to be moved 
less than 200 feet; carts, wagons, and wheeled scrapers when the 
material is to be moved less than 600 feet; cars are employed 
for moving material more than 600 feet, especially if the amount 

is large. 


CIVIL ENGINEERING. 


496 


Excavations.—In making deep excavations it is frequently 
advisable to vary the normal inclination of the side slopes of cuts 
to meet local conditions. If very hard material is encountered, 
as gravelly soil, the inclination may be increased; if, on the con¬ 
trary, loose sand or soft clayey earth, which retains the water 
and becomes greasy, is encountered, the inclination must be re¬ 
duced. The last material has a natural slope, when wet, of 
about one-third. If it is necessary either on account of the cost 
or on account of encroachment on private property to maintain 
the material at a steeper slope than its natural slope, the side 
slope of a deep cut may be terraced by leaving horizontal benches 
a few feet wide at vertical intervals of 10 or 12 feet; this will 
usually localize the sliding of the material of the slopes. The slopes 
may also be revetted or supported by retaining-walls on either 
side of the carriageway. 

If the material of the cuts is very hard, it is loosened with 
explosives. 

In making excavations care is taken not to disturb the line 
and grade stakes; these are usually left on small pyramids of 
earth which are not removed until the cut is brought to grade 
and tested. 

Embankments.—An embankment is usually formed by 
hauling the material from the nearest excavation or cut and 
dumping it so that the beginning of the embankment is of full 
height and depth. The embankment is then prolonged by dump¬ 
ing the material at the end of the completed part. As an em¬ 
bankment thus made will settle, it will be made higher than 
the grade given on the profile sheet. The height of the fill at 
each stake is indicated to the workman by erecting poles with 
cross-pieces at the proper height. 

If the embankment is made over marshy ground, it is sup¬ 
ported by a floor made of rows of sleepers or longitudinal pieces, 
which break joints and are covered with a platform of boards, 
or with layers of fascines covered with brush. The platform 
prevents the unequal settling of the roadway, and the brush 
prevents the earth from sifting through the fascines. The 
boards and fascines may also be covered with straw or similar 
material. 

Road embankments over marshes must be made high enough 


HIGH IV.AYS. 


497 


to keep the road surface about a foot or two above the surface 
of the water when the latter is at its highest stage. 


Drainage. 

The system of drainage of a road comprises the culverts which 
carry underneath the roadway the continuous and intermittent 
streams which intersect the line of the road; the side ditches which 
carry the water which falls on the road, as well as that which flows 
towards it from the adjoining lands, to the culverts or to natural 
lines of drainage which will carry it away from the road; the sub¬ 
soil drains which carry off the ground-water and keep the road¬ 
bed dry; the water-breaks which on slopes turn the water from 
the carriageway into the gutter. 

Culverts.—The term culvert is limited to spans of about 
12 feet; when the span exceeds 12 feet the structure is a bridge. 
Theoretically the area of cross-section of a culvert should be 
just sufficient to carry off the water which reaches it in the heaviest 
rain and when its drainage area is in the best condition to dis¬ 
charge this water rapidly; no damage will then be done to the 
road or to the adjoining property. 

The area of the culvert opening will depend, therefore, on the 
maximum rate of rainfall, the area of drainage basin above the 
culvert, the slope of the basin, the condition of the soil as affect¬ 
ing the amount of water which it will absorb in a unit of time, 
and the condition of its surface as interfering with the free flow 
of the water. 

It is evident that the accurate solution of the problem is a very 
difficult one; for all ordinary cases reliance must be placed upon 
empirical formulas modified by good judgment. The one com¬ 
monly employed (Myer’s) is 

Area of cross-section in square feet =cv drainage area in acres, 

in which c is a coefficient whose value depends on the character 
of the surface; it is 1 for a flat country, 1.6 for a rolling country, 
and as high as 4 for a mountainous country. The flood height 
and the area of cross-section of a stream in time of flood can 
often be best determined by making inquiries of people living 



49 s 


CiVIL ENGINEERING. 


near it, or by a careful inspection of its banks and the culverts 
or bridges along its course. 

For a very important culvert, whose failure to carry off the 
water promptly might cause serious damage, the amount of 
water which will probably reach the culvert is determined by 
some formula of the following form: 

> (Z 

in which D = cubic feet per second per acre reaching the culvert; 

c = coefficient varying with amount of obstruction to 
flow from 0.75 to 0.30; (The first value corre¬ 
sponds to the drainage of paved city streets, and 
the latter to the drainage of village gardens and 
macadam streets. Frozen meadow-land covered 
with a thin film of ice would have a coefficient of 
about 0.60.) 

r — maximum rainfall in cubic feet per second per acre, 
which is approximately equivalent to the rainfall 
in inches per hour; 

s = slope of ground in feet per thousand feet; 
a = number of acres in the drainage area. 

To apply this formula (Biirkli-Ziegler) a contoured map of 
the drainage area is necessary. 

Culverts should always be made in solid ground at the base 
or sides of a fill, and never in the made ground of the fill. If 
constructed in made ground, they will be broken when the em¬ 
bankment settles. 

Small culverts up to 2 feet span are usually made of glazed 
sewer- or cast-iron water-pipe, 12 to 24 inches in diameter. The 
iron pipe is best when the top of the culvert must be close to the 
surface of an unpaved road. To increase the capacity of the 
culvert two or more of the pipes may be laid side by side. 

To construct a pipe culvert a trench is dug in the solid ground, 
and in this the pipe is laid with a uniform fall at least sufficient 
to drain the pipe. The joints are carefully calked with cement 
or clay, and the trench is filled with earth carefully rammed. 
The ends of the culvert should if possible be supported by im- 


HIGH IV.AYS. 


499 


bedding them in low retaining-walls built parallel to the roadway. 
If the foundation is not solid ground, the soft material should be 
excavated and filled with good earth well tamped, so that the 
water will not undermine the pipe. If there is any danger of 
settlement of the middle part of the culvert, this should be pro¬ 
vided for by increasing the fall near the outlet. 

Medium-sized culverts are usually rectangular or box cul¬ 
verts or arch culverts. The box culvert is made of dry or cemented 
stone walls, and is covered with slabs of stone. As the resistance 
of stone to a bending force is small, the cover is also sometimes 
made of two or more layers of slabs in which each layer projects 
slightly beyond the one beneath it, thus gradually decreasing the 
span. The bottom of the culvert is paved with stone, preferably 
laid in mortar. 

When more durable material is not av ilable box culverts 
are also made of beams and planks or of square timbers. When 
the span exceeds 4 feet there is usually a central vertical parti¬ 
tion which divides the culvert into two equal channels, and 
reduces the thickness of the covering timbers. 

Arch culverts of stone, brick, and concrete are employed for 
spans of from 2 to 12 feet. The form of the arch is either the full 
center or the segmental. The depth of the arch at the crown 
may be determined by the formula d = 1 foot+ ^5, in which 
d = depth and 5 = span in feet. In arches whose span does not 
exceed 3 feet the thickness of the arch may be uniform through¬ 
out ; as the span increases the thickness of the arch at the spring- 
ing-lines increases more rapidly than at the crown, and in arches 
whose span exceeds 6 feet the thickness at the springing-lines is 
double the crown thickness. The thickness of the abutments for 
spans up to 12 feet may be deduced from the empirical formula 
/ = 3 feet + jo ^ + tV ( s ~3 feet), in which t is the thickness and h 
is the height of the abutment, and 5 is the span of a full-center 
arch in feet. The value of t is increased £ for a segmental 
arch. 

The box and arch culverts may extend the full width of the 
carriageway and their side slopes, and terminate in a vertical 
wall, or they may extend only the width of the carriageway. In 
the latter case the abutment-walls are prolonged as wing-walls 
through the side slopes and gradually contract the waterway. 


5 °° 


CIVIL ENGINEERING. 


Spandrel- or head-walls confine the earth filling on the top of the 
culvert proper. The paving of the bottom of the culvert is 
extended to include the space between the wing-walls. 

If the culvert is a wide one, the pavement of the bottom may 
be made slightly concave upwards. 

Side Ditches.—The side ditches are usually trapezoidal in 
cross-section and should have sufficient area of cross-section to 
carry the drainage during the heaviest rainfall when the surface 
is in the best condition to allow its rapid drainage. The ditches 
are paved only when the grade of the road is such that they would 
otherwise be rapidly eroded. When the road is constructed at 
or near the foot of a long slope one or more additional ditches 
should be dug along the face of the slope, parallel to the road, 
to catch the surface drainage before it reaches the side slopes 
of the road. This will prevent the washing of these slopes in 
heavy rains. 

Deep side ditches assist in draining the subsoil of the road, 
and are especially valuable when no subsoil system of drainage 
is provided. 

Subsoil Drains.—A subsoil drain is a covered channel de¬ 
signed to remove the water which penetrates the road-bed, and 
thus secure a dry foundation for the wearing surface. The drain 
must have an unobstructed channel inclosed by an envelope 
through which the water can percolate to reach the channel. 
A drain made of porous terra-cotta pipes called drain-tiles, or a 
small box culvert constructed of dry stones, will fulfill these re¬ 
quirements. 

If the carriageway is not covered by an impervious coating, 
some of the water which falls on the surface or results from the 
melting snow will penetrate the road-bed. The percentage ab¬ 
sorbed will depend on the form of cross-section and the ruts and 
depressions in the surface. 

If the carriageway is lower than the adjoining land, some of 
the water falling on this land will also reach the road-bed if an 
impermeable stratum underlies the road-bed. The amount 
which reaches it in this manner depends on the position and dip 
of the impermeable strata. 

If the road-bed is sandy, the water which reaches it will be 
rapidly carried away through the soil and will do no injury; if ? 


HIGHWAYS. 


5 QI 

however, the road-bed is clayey, the water is retained and the 
road-bed becomes very soft. 

Subsoil drains are therefore most necessary in deep cuts in 
clayey soil, on roads which have a natural earth surface. They 
are usually unnecessary in embankments and in sandy soil. 

The best form of drain is a line of porous-tile pipe, usually 
about 4 or 5 inches in diameter, laid on a firm bottom, true 
to grade, and covered with a layer of broken stone or gravel a 
foot or more in thickness. As the gravel or broken stone will 
retain the silt, the pipes may be laid without collars over the 
joints. If no layer of broken stone is employed, the joints must 
be covered with tile or clay collars to prevent their obstruction 
by silt. 

The tiles are usually laid in a trench about 3 feet deep, but 
this depth is varied to meet local conditions. 

The position of the drains in the road depends on the direc¬ 
tion from which the water comes. For surface drainage there 
should be one deep line bisecting the carriageway longitudinally, 
or two shallower lines trisecting it. When the water comes 
from one side only, as on a macadamized road along a hillside* 



the drain should be on that side, with branches if necessary lead¬ 
ing to it from the carriageway. If the water comes from the two 
sides, as on a macadamized road in a cut, a drain should be con¬ 
structed along each side of the road under the side ditches, as 
these are liable to be filled by the caving of the slopes. 

Water-breaks.—Water-breaks are shallow ditches or low 
embankments which are constructed diagonally across the car¬ 
riageway on steep grades, to prevent the water from running in 
the wheel-ruts and thus eroding the surface. They are unneces¬ 
sary on paved carriageways or on macadamized carriageways 
whose surfaces are properly crowned. Small cross-drains, cov¬ 
ered by a grating and connected with the sewers, are sometimes 







502 


CIVIL ENGINEERING. 


constructed across paved roadways on long grades to prevent 
the accumulation of too much water at the foot of the slope. 

Catch-basins.—A catch-basin is a shallow well constructed near 
the curb of a street; it is designed to catch the water which flows 
from the gutters, and remove its silt before discharging it into the 
sewers. The details of construction are given under “ Sewerage.’’ 

Earth, Gravel, and Macadam Roads. 

The top or wearing surface of the carriageway should be 
impervious to water , incompressible , offer little resistance to the 
movement of a vehicle , and furnish a good foothold to the horse. 

It should be added that the surface should not become 
dusty. If the highway is an unpaved one, this can be pre¬ 
vented in hot, dry weather only by sprinkling daily with water, 
or occasionally with crude oil. For country roads the latter is 
the only practicable method. 

Earth Roads.—An earth road is one in which the wearing 
surface of the carriageway is the earth excavated from the sides 
of the carriageway or from the ditches. It fulfills none of the 
requisites of a good wearing surface, unless well drained and 
kept in thorough repair. Even then its tractive resistance is 
at least ioo pounds per ton. The quality of the wearing surface 
depends upon its composition; it is best when composed of sand 
or gravel with just sufficient clay or loam to bind it thoroughly; 
it is worst when the sand or the clay alone is used. 

If the cross-section is of the general form shown in Fig. 160, 
the carriageway proper should have a crown of and the 
side ditches should be deep enough to assist in draining the 
road-bed. As this form of cross-section necessitates the excava¬ 
tion of the ditches by hand, a more common form is an arc of 
a circle connecting the bottoms of the side ditches. A high 
crown is necessary to prevent the surface-water from sinking into 
the road-bed, when the road is worn by travel. 

The road covering being compressible is easily worn into 
ruts which hold the water, and in clayey soil render the road 
impassable in wet weather unless the road is provided with 
subsoil drains. The compressibility of the surface may be 
reduced by rolling it with a heavy roller and by making the road 
covering of a mixture of clay with sand or gravel. The surface 


HIGHIVA YS. 


5°3 


drainage can be kept intact only by filling the ruts with good 
material as soon as made. Large stones should never be used in 
making these repairs. 

As dependence must be placed largely upon heat and free 
circulation of air to dry the road-bed, a road in clayey soil should 
never be shaded by overhanging trees; a sandy road, being better 
in a moist than in a dry condition, should have as much shade as 
possible. 

Besides the grading machinery employed in the construction 
of the road-bed as above described, the principal machines em¬ 
ployed in its improvement are the simple and elevating scraping 
graders. The simple scraping grader is a four-wheeled vehicle 
which supports a heavy steel blade, either plane or curved, which 
may be set so that its cutting edge is oblique both to the axis of 
the road and to the horizontal plane. As it is pulled along it 
cuts the surface to an inclined plane and moves it from the sides 
toward the center. The carriageway may thus be given a rounded 
surface. The elevating scraping grader has, in addition, a 
carrier by means of which the earth excavated from the side 
ditches is conveyed directly to the middle of the carriageway. 
Both machines may be employed in constructing new roads and 
in repairing old ones. 

Gravel Roads.—In a gravel road the wearing surface is a 
bed of compressed gravel usually from 8 to 12 inches in thickness. 
When in thorough repair the wearing surface is all that can 
be desired, but it wears away rapidly under heavy travel and 
requires a thoroughly drained road-bed. The quality of the 
wearing surface depends on the character of the gravel and 
binder. The gravel should vary in size from 2 inches down¬ 
wards so as to reduce the volume of voids to a minimum, and 
should be united by a binding material which will not absorb 
water. The cement formed from the crushed gravel is the best 
binder, but it will take some time before a surface of loose gravel, 
such as is dredged from a river bottom, can be rendered com¬ 
pact. Clay loam, just sufficient to fill the voids, makes a good 
binder, but if in excess, will cause the carriageway to become 
soft and muddy in wet weather. A small quantity of lime or 
oxide of iron mixed with the clay improves its binding qualities. 

The crown of a gravel road may be made somewhat less than 


CIVIL ENGINEERING. 


5°4 

an earth road, or about to 3 1 -. The form of cross-section is 
like that of a macadam road. 

The gravel may be taken from pits or river-beds. The first 
has usually an excess of sand or clay and must be screened; of 
the two materials in excess the sand is least objectionable. River- 
gravel packs slowly and is usually mixed with clay or loam to 
make it bind more readily; the mixture is, however, less durable 
than the pure gravel. 

In making gravel roads the carriageway is excavated to sub¬ 
grade and rolled; if the ground is clayey, the road-bed is drained 
or the gravel is placed on a bed of coarse broken stone. A layer 
of gravel about 4 inches thick is spread on the subgrade, moist¬ 
ened, and rolled until it is packed. The other layers are dis¬ 
tributed and rolled in the same manner. In contract work the 
amount of gravel is determined by running lines or levels over 
the rolled subgrade and the finished work. The wearing surface 
may be made of uniform thickness, or the thickness of the sides 
of the carriageway may be made 2 or 3 inches less than the thick¬ 
ness at the center. 

Macadam.—The term macadam is applied to stone brokea 
into small fragments to be used in making broken-stone roads, 
and the terms macadam road and macadam pavement , applied 
to highways thus improved, are derived from John Macadam and 
his son Sir James Macadam, who were employed in improving 
the highways of Great Britain during the first half of the nine¬ 
teenth century. 

The proper way to construct a macadam road was described 
by Sir James Macadam about 1850 as follows: 

“The road-bed is thoroughly drained and all large stones 
are carefully removed. The surface of the road is then graded 
to a lateral slope of 1/36. As sand, sandy earth, or any other soft 
and dry material makes the best foundation for a road, the broken 
stone is placed directly on the ground. On first-class roads there 
are three layers of stone. The first consists of any sound stone 
which is broken into fragments weighing not over 6 ounces and 
screened to remove all dirt. This layer is 4 inches thick and is 
compressed by travel or with a roller. The second layer, 2 or 
3 inches thick, is of stone similar to the first, and is compressed 
in the same manner. The third layer is of granite or other hard 


HIGHWAYS. 


505 


stone, and its fragments are of the same size as those of the other 
layers. This is the wearing surface proper and is 3 inches thick. 
No sand or binding material should be spread over the surface 
to fill the interstices, but these should be filled by the fragments 
and dust produced by the travel on the road. The repairs should 
consist in picking the road surface to a depth of 1 inch and then 
placing on a new layer. A second-class road needs only a 4-inch 
base and a 3-inch wearing surface, and a third-class road only 
a 3-inch base and a 2-inch surface.” 

Modern practice differs from the method prescribed in only 
two essential particulars. 

It is customary to excavate a trench to hold the broken stone. 
The bottom of the trench or the subgrade is rolled with the 
heaviest roller obtainable and all its soft places filled with sandy 
earth or ashes. The subgrade when thoroughly rolled should 
be parallel to the finished surface of the carriageway, and at 
the proper distance below it. The thickness of the macadam 
depends upon the character of the soil, the amount of travel, 
and the frequency of repairs. The standard depth under favor¬ 
able conditions is 6 inches; this may be decreased to 4 inches 
under very favorable conditions, and is increased to as much as 
12 inches if the conditions are more or less unfavorable. 

Binder.—The Macadams insisted that the surface of the road 
must be impervious, but at the same time insisted that no bind¬ 
ing material should be used to close the crevices between the 
stones. In the course of time these crevices will naturally be 
filled by the fragments and dust produced by travel; in the mean¬ 
time, however, the surface is rough and permeable. It is now 
customary to fill the crevices at once by placing some suitable 
binding material on the surface layer after it is thoroughly rolled, 
and to force it into the crevices by additional rolling. This 
gives a smooth, impermeable surface at once. 

The natural tendency of inexperienced road-constructors is to 
employ clay or some form of clayey earth for the binder because 
of its adhesive properties. It is, however, the worst material 
which can be used for this purpose, because it holds the water, 
which settles in the depressions of the road, and becomes soft 
and greasy. Sand or sandy gravel is often used and forms a good 
binder, especially if the wearing layer is limestone. Stone screen- 


5° 6 


CIVIL ENGINEERING. 


ings from the crusher make the best binder; those from trap-rock 
give the best results. As a rule, a silicious binder gives the best 
results on a limestone macadam, and a calcareous one on a 
silicious rock macadam. If circumstances require the use of clay, 
it should be used sparingly; one containing iron is the best. 

The binder is usually placed only on the wearing layer, 
though some engineers place it on the foundation layers as well. 
This last method probably gives a result similar to that obtained 
by Macadam when he had each layer compressed separately by 
travel. With a perfect binder the result is approximately a con¬ 
crete, and the pavement is ideal; if, however, the binder retains 
the water and softens, it will assist in destroying the pavement. 
If only a surface binder is employed, the water which passes through 
it will at once reach the subgrade. If this complies with Mac¬ 
adam’s requirements, no damage will be done, as the water will 
be at once drained away; if, however, the soil retains the water, 
it will soften, be pressed up into the macadam and assist in the 
destruction of the pavement. When the soil is dry and the 
binder is good, a layer of binder worked into each layer of mac¬ 
adam will improve the pavement, otherwise it is inadvisable and 
uselessly expensive. 

Stone.—As to its characteristics, the stone in the surface 
layer should be tough, hard, of uniform strength in all directions, 
its dust should have cementitious qualities, and it should resist 
the destructive action of the weather. The igneous or unstratified 
rocks, such as porphyry, trap, basalt, granite, syenite, and diorite 
have the best average qualities. The limestones, the meta- 
morphic rocks, quartzite, gneiss, schist, and millstone grit, rank 
second. In the igneous and metamorphic rocks sodium is an ele¬ 
ment of weakness, as it usually mak the stone more perishable. 
In the second class the difference between the best and poorest 
of any variety is greater than in the first class; greater care is there¬ 
fore necessary in selecting the stone. In the foundation layers 
any sound stone may be used. Sandstones are often used in 
these layers. 

Rolling.—The subgrade, the successive layers of stone, and 
the binder are all rolled separately with a heavy roller. If a 
horse-roller is used, its weight should be about 5 tons, and if a 
steam-roller is used, its weight should be about 15 tons. A heavy 


HIGHWAYS . 


5°7 


roller will make the road-bed more unyielding and the pavement 
more compact than a light one. The layers of stone and the 
binder are sprinkled with water before they are rolled. 

Telford Macadam. —Telford-macadam pavement is one in 
which flat stones placed on edge replace the foundation-layers 
of ordinary macadam. It is named from Thomas Telford, a 



Fig. 167 


celebrated Scotch engineer who was a contemporary of the elder 
Macadam, and was also employed on highway improvement in 
Great Britain (Fig. 167). 

This method of construction was first employed in France. 
The highways in France were, before 1764, repaired only twice a 
year, and then by unskilled laborers; they were therefore made 
in a very substantial manner. At the base were two or three 
layers of flat stones laid on their sides; these were covered with 
broken stone to such a depth that the whole pavement was 18 
inches thick in the center and 12 inches on the side. 

After 1764 the highways were regularly maintained, and 
in a memoir published in 1775 the French engineer Tresaguet 
described a method of construction by which the depth of the 
pavement could be greatly reduced. He laid the foundation- 
stones on edge, in rows, across the carriageway; the height of 
this layer was 7 inches. Its crevices were filled by fragments 
of stone driven in with a hammer. The top layer consisted of 
3 inches of broken stone of ordinary dimensions. The road¬ 
bed was usually excavated to a subgrade parallel to the finished 
surface, and a row of stones on edge, 10 inches high, was placed 
on either side to hold the broken stone in place. These stones 
formed a curb which did not project above the surface of the 
carriageway. He made his carriageways both convex and con¬ 
cave in cross-section. 

Telford varied this method of construction bv making the 
subgrade horizontal and securing the crown by varying the depth 
of the foundation-stones; this was, however, not an essential 
detail. The foundation-stones were from 3 to 5 inches thick 
on top, and from 8 to 14 inches long. He omitted the border- 
stones and usually covered the shoulders of the pavement with 






5°8 


CIVIL ENGINEERING. 


a layer of gravel. As he was engaged only with the improve¬ 
ment of important roads, he made the surface layer of broken 
stone 6 inches thick. In a road 30 feet wide he first adopted 
a crown of 9 inches, but later considered 6 inches sufficient. His 
roads were always well drained and carefully constructed. Like 
macadam pavements, the foundation-course of a telford pave¬ 
ment may be made of sound stones which are either too soft or 
too brittle to be used in the surface layer 

Comparison of Telford and Macadam Methods.—The tel¬ 
ford foundation is more open than the macadam; the water, 
which reaches the road bed either from above or below, can 
thus escape if the bed has any longitudinal slope. The inter¬ 
stices of this foundation are open, and hence the ground-water 
cannot reach the binder of the top surface to soften it. In the 
macadam foundation the interstices are usually filled with binder 
or dust, which prevents the escape of the water and allows it to 
be drawn up and distributed through the binder by capillary 
attraction. This softens the pavement. Macadam pavement, 
being the cheaper, should therefore be laid when the soil is sandy 
and allows the escape of water through it; if laid in wet places, 
the foundation-layer should be of stones 3 or 4 inches in diam¬ 
eter, and the binder should be confined to the top layer. The 
foundation of the telford pavement must be laid by hand; it is 
therefore more expensive. The telford pavement is less liable 
to be affected by ground-water, and is therefore best when a pave¬ 
ment must be laid on clayey soil, especially in cuts where the 
water is liable to accumulate. 

Block and Sheet Pavements. 

The macadam and telford pavements are not suitable for 
city streets because of the dust and mud, the necessity of constant 
repairs, and the impossibility of keeping the surface clean with¬ 
out injuring it. For these reasons sheet and block pavements 
are laid on these streets. 

An ideal city pavement is smooth without being slippery , 
elastic without losing its shape , impervious to liquids , and con¬ 
tains no decaying substances. On such a pavement the tractive 
resistance of a vehicle and the noise produced by the travel is 
reduced to a minimum, and the pavement contributes to the health 


HIGH IVAYS. 


5°9 


of the community. In addition to the above qualities, the cost 
oj construction and maintenance should not be excessive. 

A city pavement usually consists of a wearing surjace which 
is exposed to the concussion and friction of travel and must be 
renewed from time to time,‘and a foundation which supports the 
wearing surface and prevents it from losing its shape. 

Wearing Surfaces.—The ordinary wearing surfaces are made 
of blocks of wood, natural and artificial stone, and asphalt con¬ 
crete, and of sheets of asphalt mortar. 

All block pavements are laid in a similar manner, with the 
blocks in rows as nearly as possible perpendicular to the line 
of travel. Between street intersections the rows are perpendicu¬ 
lar to the axis of the street; at the street intersections the rows 
may be parallel to the diagonals which bisect the four angles 
made by the two axes. The blocks break joints in the direction 
of the travel. The blocks are placed by hand on a cushion of 
sand, and are brought to a firm bearing by tamping the blocks 
with a heavy hammer or by rolling the pavement with a light 
roller. A steel shoe covering several blocks is placed on the 
pavement to receive the blows of the hammer; if the blocks are 
brittle, a wooden shoe is employed instead of a steel one. Tamp¬ 
ing, although more expensive, is usually preferred to rolling 
because the blocks are not displaced laterally. The thickness 
of the sand cushion varies from \ inch to 2 inches, depending 
on the allowable variation in the depth of the blocks and the 
smoothness of the surface of the foundation. 

Foundations.—The best foundation for all forms of pave¬ 
ment is a bed of concrete whose upper surface is fairly smooth 
and parallel to the finished surface of the pavement. The bed 
is of uniform thickness and is laid on the subgrade, which has 
been graded so as to be parallel to the finished surface and then 
thoroughly rolled with a heavy roller. The form of cross-section 
of the carriageway is determined by making the height of the 
crown equal to or of the width of the roadway and the height 
of the quarter-points j the height of the middle point. The sub¬ 
grade and the surface of the concrete are tested by measuring 
down from a horizontal cord connecting the curbs or attached 
to stakes driven on either side of the carriageway. 

A concrete foundation is absolutely rigid, and will hold the 


CIVIL ENGINEERING. 


5 IQ 

wearing surface in position at every point, and thus prevent 
the formation of local depressions which increase the concussion 
and friction at those points and hasten the destruction of the 
wearing surface. If the foundation is removed for the purpose 
of laying water-mains or sewers, or making house connec¬ 
tions, it is easily restored, and the pavement will show no de¬ 
pression. 

The normal thickness of the concrete foundation is 6 inches . 
This may be reduced to 4 or 5 inches if the concrete is rich, care¬ 
fully prepared, and laid upon a thoroughly rolled subgrade, 
and the foundation is not liable to be disturbed for the purpose 
of laying sewers, mains, etc. The proportion of the ingredients 
is usually 1 natural cement, 2 sand, and 4 stone, or 1 portland 
cement, 3 sand, and 6 stone. If the cement is carefully tested and 
the mixing carefully inspected, the proportion of cement may be 
reduced; in the city of Washington the proportions are: 1 port- 
land cement, 4 sand, 5 gravel and 5 stone. In place of gravel 
the run of the crusher may be substituted; it differs from the 
other stone in containing material smaller than J-inch pieces. 

Where concrete is considered too expensive a bed of gravel 
or cinders of about the same thickness and compressed with a 
heavy roller may be employed. As this form of bed will nearly 
always show depressions over cuts made for the purposes above 
described, it will probably be more expensive in the end than 
a concrete base, because of the more rapid destruction of the 
wearing surface. 

Wood Pavements. — A pavement of untreated soft-wood 
blocks on a rigid foundation is the most noiseless and elastic 
and the least slippery form of pavement, but is unsanitary and 
perishable on account of its absorption of liquids, and wears 
rapidly under heavy travel. As the blocks will decay in three 
to five years, depending on the care taken to keep them clean, 
and will, except under the heaviest city travel, stand the wear of 
travel for a much longer time, it is customary to preserve the blocks, 
by the creosote process. This makes the blocks impervious and 
protects them from rot, but makes them less elastic and more 
slippery. On the great thoroughfares of London a pavement 
of this kind is worn out in five to seven years, but under light 
travel it would probably last fifteen years or more. A first-class 


HIGH WAYS. 5 ii* 

pavement of creosoted Georgia pine costs in this country about 
$3. 2 5 per square yard. 

If hard-wood blocks are used the pavement is more slippery 
but less perishable than one of soft wood. In Australia there are 
pavements of very hard, dense, untreated wood that have stood 
ordinary wear for fifteen years; these pavements are, however, 
extremely slippery. 

The best wood pavements consist of soft creosoted or dense 
hard-wood blocks whose surface is 3" X9". The maximum 
depth of the blocks is 6 inches, and the minimum 4 inches. 

The best creosoted blocks are fir and yellow pine; the best 
natural hard-wood blocks found in the market are the Jarri and 
Karri woods of Australia. 

Hard or creosoted soft-wood blocks are extensively used in 
paving the streets of London and Paris, because they are less 
noisy and less slippery than the stone-block and rock-asphalt 
pavements, which are the other principal forms; in these cities 
wood pavements are laid on the streets which are subjected to 
very heavy travel. In this country soft untreated wood blocks 
were laid in nearly all the principal cities between 1870 and 1875. 
This pavement, called the Nicholson, was usually laid with wide 
transversal joints filled with gravel, and on a wooden platform 
consisting of two layers of pine boards coated with tar. The 
surface was covered with tar and gravel. These pavements were 
short-lived and led to a distrust of wood as a paving material. 
The Nicholson pavement was followed by the cylindrical cedar- 
block pavement laid extensively in the northern central States. 
This was an improvement, but was also perishable and unsatis¬ 
factory. The introduction and development of the Trinidad 
asphalt pavements between 1875 and 1880 furnished a suitable 
substitute which was generally adopted in cities of the United 
States. It is only since about 1890 that good wood pavements 
have been laid in this country. 

A first-class wood pavement has a foundation of 6 inches of 
concrete. If the concrete is finished off with a smooth surface 
by covering it with a thin layer of mortar, the blocks may be set 
on this mortar; as a rule, however, the concrete is covered by 
an inch of sand or a layer of asphalt. 

The blocks are rectangular prisms with the grain of the wood 


5 12 


CIVIL ENGINEERING. 


vertical. If thoroughly creosoted, so that they will not absorb 
moisture and expand, they may be laid with close joints in both 
directions; as a rule the blocks are laid in close contact in each 
row, and the rows are separated by a space J to f inch by the inser¬ 
tion of thin strips which are afterwards removed and the joints are 
filled with cement grout, or by asphalt cement covered by an inch 
of grout, as the latter stands the wear of travel better than the 
asphalt. An expansion-joint is left at the gutter ends of each 
row, and the joints between the rows are filled after the blocks 
have expanded. Close joints are found to resist heavy travel 
better than open ones. A wood pavement must be kept in 
repair by the removal of the blocks which are found to be 
defective; otherwise depressions are formed which hasten its 
destruction. 

Plank and corduroy pavements are crude forms of wood 
pavements; they are employed to carry a road across a swamp 
or over dry sand into which the wheels would sink to a considerable 
depth were the wagon not supported by some form of platform. 
The foundation of the platform is two or more rows of longitudinal 
sleepers, usually logs. In the plank pavement these are covered 
with a floor of 2- or 3-inch plank, and in the corduroy by sap¬ 
lings placed in close contact and held in place by wheel-guards. 
The upper surfaces of the saplings may be roughly hewn, or 
they may be covered with brush or grass and a layer of earth. 

Stone-block Pavement.—A pavement of granite blocks on 
a rigid foundation is the most durable of all pavements, and 
requires no attention for many years after it has been laid; it 
may be made impervious by filling the joints with hot gravel 
and asphalt cement pitch. As the surface is unelastic and 
rough, the noise made by the passing travel is so great that in 
many cities they are being gradually replaced by the less noisy 
forms. 

The blocks are made of granite, trap, and sandstone. The 
granite blocks are best, since they are more durable than sand¬ 
stone and less slippery than trap. The sandstone must be one 
with silicious or ferruginous binding material. 

The size of the blocks varies considerably in different cities. 
In Liverpool, which has probably the best stone-block pavements, 
the blocks are about 3J inches square and 6 inches deep; they 


HIGH W.A YS. 


5*3 


must be culled so that the maximum variation shall not exceed 
J inch. These blocks are laid on a half-inch cushion of gravel 
on a 6-inch concrete base. In Washington the blocks are 6 
to 8 inches long, 3 to 4 inches wide, and 5J to 6 inches deep; these 
blocks are laid with close joints filled with pitch on a 2-inch 
sand cushion. The foundation is 4 inches of rolled gravel. In 
other large cities the blocks are 8 to 12 inches long, 3J to 4J inches 
wide, and 7 to 8 inches deep; as a rule these are not laid on con¬ 
crete. If laid on a concrete foundation, small blocks are as un¬ 
yielding as large ones and are less easily displaced; if the founda¬ 
tion is not rigid, the blocks must have a larger bearing surface. 
The size of the blocks is less important than the uniformity in 
length and in breadth, since the latter governs the dimensions 
of the joints, which should be as narrow as possible. 

The joints in the best pavements are filled with hot gravel 
and asphalt cement, tar, or pitch; those of inferior pavements 
are filled with gravel only. 

The price of stone blocks at the quarry is about $1.15 per 
square yard, and the cost of laying the blocks on a 6-inch con¬ 
crete base after they have been delivered in the city is about 
S1.10 per square yard. To the total cost of $2.25 per square yard 
must be added the freight charges and the contractor’s profits. 

Pavements with a stone wearing surface were introduced into 
Europe by the Romans in the construction of their military roads 
and in the paving of their city streets. These pavements con¬ 
sisted of irregular-shaped slabs, usually of large area, which were 
supported by a firm foundation of flat stones covered by a layer 
of concrete. In many Italian cities the influence of the Roman 
method of construction is seen in pavements whose surface is 
made of large flat stones. In the principal street of Genoa the 
blocks are nearly 2J feet long and 1 foot wide; grooves are cut 
in the stones dividing them into narrow strips. 

The cobblestone pavement is a rough form of block pavement 
which is made of the spherical or egg-shaped stones found in the 
beds of streams and in fields. It is very rough, very slippery, hard 
to keep clean, and destructive to vehicles. The noise of travel 
is much greater than over stone blocks. The best of these pave¬ 
ments is made of stones 4 to 6 inches long, 2 to 4 inches in diam¬ 
eter. The foundation is a bed of compressed gravel 4 or 5 inches 


5*4 


CIVIL ENGINEERING. 


thick and a cushion of sand. The stones are laid with close 
joints, covered with sand, and then tamped with a heavy rammer 
until they are firmly bedded. 

Brick Pavements.—A brick pavement is simply a stone-block 
pavement in which blocks of natural stone are replaced by blocks 
of artificial stone. Bricks are somewhat less durable than stone 
blocks, but they can be made of uniform size and with smooth 
faces; the resulting pavement has therefore narrower joints and 
a smoother surface, both of which greatly reduce the noise of 
travel and the resistance to traction. The pavement is also 
cheaper than stone, since the cost of the brick at the kiln is about 
one-half the cost of stone blocks at the quarry. 

The bricks used for pavements are made of selected clay, 
usually shale, and are burned at a high heat just short of vitri- 
faction. They are called vitrified bricks if of the same size as 
ordinary bricks, and vitrified blocks if of larger size. Blocks 
are more commonly used than bricks. 

The blocks are usually 3 inches wide, 9 inches long, and 4 
inches deep. If made of shale, they are reddish brown in color, 
and if of fire-clay, they are mottled yellow. To make a durable 
pavement the blocks must be tough and thoroughly burned; to 
make a smooth pavement they must be of regular form and of 
uniform size. The edges of the blocks are often rounded, and the 
vertical faces are grooved or studded. The rounded edges 
reduce the loss of weight in the rattler test to be described, but add 
nothing to the value of the blocks; the studs and grooves are 
to make space for the grout or asphalt cement with which the 
joints are filled, but this is unnecessary. A brick with square 
edges and plane faces is probably the best form. 

When bricks are compared with a standard or with each other 
they are usually subjected to two tests, the absorption and the 
rattler test. The absorption test consists in determining the per¬ 
centage of water by weight which will be absorbed in forty-eight 
hours by a block which has been thoroughly dried in an oven. 
As the surface can be easily made impervious by a suitable coating, 
the blocks are broken into two nearly equal pieces. A good 
paving-block should absorb very little water. The rattler test 
is made by placing the paving-blocks and small cast-iron blocks 
in a rattler and rotating the rattler at a fixed speed for a definite 


HIGH WAYS. 


-515 

time. The toughness and durability of any specimen is assumed 
to vary inversely with its percentage loss of weight in the rattler. 
A rattler is a metal barrel placed in a horizontal position and 
rotated by machinery about its longer axis. It is employed in 
cleaning castings. In the rattler the edges of the bricks are worn 
away by concussion and friction, and the bricks, if too soft, are 
broken. Standard testing rattlers are made for this, which is 
considered the most valuable of the paving-brick tests. The 
blocks are also sometimes tested for compressive strength, but 
this is not an important test. 

In laying the pavement the bricks are placed by hand on edge 
on a cushion of sand 1 or 2 inches thick. This thickness depends 
on the smoothness of the surface of the foundation. The blocks 
are then forced to a firm bearing by dropping a heavy tamp on a 
short thick plank, or the pavement is rolled with a light roller. 
The former displaces the bricks less than the latter, but is more 
expensive. 

If the joints are filled with cement grout, it is necessary to 
make expansion-joints parallel to the axis at the curb, and joints 
perpendicular to the axis at intervals of 200 feet; these joints are 
filled with asphalt cement or pitch. In inferior pavements all 
joints are filled with sand and are not impervious. 

While a concrete foundation 4 to 6 inches thick is better than - 
any other form, the pavement may be laid on rolled macadam, 
gravel, or cinder. When these pavements were first laid in this 
country the foundation-course was a layer of bricks placed on their 
sides. 

Brick pavements were introduced into this country about 
1870, and because of their cheapness and other qualities have 
become the principal pavements of our interior cities. 

The price of re-pressed vitrified paving-blocks at the place of 
manufacture is about $0.60 per square yard; the common are 
about $0.10 less. The cost of lay ng the blocks on a 6-inch con¬ 
crete base is about $0.85 per square yard, and is less than the 
cost of laying stone blocks because of the regularity of the bricks, 
which makes it easier to break joints, and because of the smaller 
amount of cementing material required in the joints. To the 
total cost of $1.45 per square yard must be added the freight 
charges and contractor’s profits. 


CIWL ENGINEERING . 


5i6 

Asphalt Pavements. — Sheet-asphalt pavements are of two 
general classes: those made of rock asphalt or a natural mix. 
ture of limestone and asphalt, and those made of asphalt mortar 
or a mixture of asphalt cement and sand. The former having 
been found too slippery, only the latter class is laid in this country. 

The standard sheet pavement consists of a cement-concrete 
foundation 6 inches thick, a cushion or binder course of asphalt 
concrete ij inches thick, and a surface coat of asphalt mortar 
2\ inches thick. The binder is made by mixing asphalt cement 
with clean broken stone which will pass an inch and a half 
screen. The stone is heated to a high temperature in drums 
and then mixed by machinery with hot asphalt cement until 
every particle of stone is thoroughly coated. It is hauled to the 
street, spread, and rolled while still hot. The mortar is made in 
a similar manner by mixing hot sand and asphalt cement, and is 
also laid and rolled while still hot. The mortar is first compressed 
with hand-rollers, then covered with a thin layer of hydraulic 
cement or limestone dust, and finally compressed with a heavy 
steam-roller. Under the roller the original thickness of 2J inches 
is reduced to about ij inches. The cement closes the surface 
pores and gives it a gray color. 

A sheet-asphalt pavement is smooth, impervious to liquids, 
and, if of pure asphalt and sand, will not become brittle if exposed 
to continued moisture; in hot weather it is soft and noiseless, 
but offers considerable resistance to the passage of vehicles; in 
cold weather it is incompressible, hard, offers slight resistance 
to the passage of vehicles, but is slippery, especially if covered 
with mud or with a thin film of ice. It is more easily kept clean 
than any other form of pavement. 

The cost of sheet-asphalt pavement on a 6-inch concrete 
base in the city of Washington was, in 1903, $1.51 per square 
yard. This did not include the preparation of the street for 
the pavement, but included the cost of maintenance for five 
years. An allowance of about 10 cents per yard per year covers 
the cost of annual repairs and the renewal of the surface and 
binder when completely worn out. 

Asphalt binder and mortar may also be laid on a macadam, 
stone-block, or cobblestone base. 

The rock-asphalt pavement is made by reducing to a powder 


HIGH WA YS. 


517 


the natural rock, which is impregnated with bitumen, heating 
the powder to a high heat, and then spreading it over the con¬ 
crete base and rolling it in the same manner as the asphalt mortar. 
This pavement has no binder or cushion. 

Asphalt-block pavements are made of asphalt blocks laid 
in the same manner as other block pavements. The blocks 
are made by compressing in molds a hot mixture of asphalt 
cement, mineral dust, and crushed stone. In the city of Wash¬ 
ington, if laid on a 5-inch concrete foundation, the blocks are 
made 3 inches deep and either 4X12 or 5X12 inches; if laid on 
a gravel base the blocks are 5 inches deep and 4X12 inches. The 
cost of the pavement, including repairs for five years, is about 
$1.80 per square yard. 

An inferior form of sheet pavement is made of asphalt or 
tar concrete; it is not so smooth nor so durable as the sheet 
asphalt. 

Comparison of Pavements. 

Sheet asphalt made of asphalt cement and sand is the most 
satisfactory form of pavement, as its tractive resistance is small, 
vehicles move over it without noise, it is easily swept and kept 
clean, it is impervious to liquids, and can be easily and thor¬ 
oughly repaired. 

The ordinary sheet-asphalt pavement must, however, be kept 
dry , clean , and in constant repair. If these three conditions can¬ 
not be fulfilled, it should not be laid. If kept moist, the ordinary 
Trinidad asphalt pavement becomes brittle and porous. For 
this reason street-gutters are usually paved with vitrified blocks. 
If the pavement is not kept clean, it will be slippery in wet weather 
and the moist dirt will cause its deterioration. Imperfections 
in sheet-asphalt pavement are certain to be developed from 
time to time; unless the bad places are at once repaired they 
will rapidly enlarge. Bermudez asphalt is more durable in 
moist places than Trinidad. In the summer the sheet asphalt 
is soft and its tractive resistance is great, but the horse secures 
a good foothold; in winter the pavement is hard and its trac¬ 
tive resistance is small, but the horse cannot get a good 
foothold. 


CIVIL ENGINEERING . 


518 

Wood pavement of creosoted soft or hard wood is also as 
smooth as sheet asphalt, less noisy in cold weather, and does 
not require such constant repairs. It is, however, not so imper¬ 
vious, nor can it be so thoroughly cleaned. If kept clean, it is 
less slippery than sheet asphalt and gives a better foothold to 
the horse in cold weather, but the contrary is true in warm 
weather. 

Asphalt-block pavement is neither so smooth nor so noiseless 
as sheet asphalt or wood, but not much inferior to them. Like 
sheet pavement it must be kept clean, otherwise the cementing 
material deteriorates. It affords a better foothold for the horse 
than sheet asphalt, and is therefore often laid on grades con¬ 
sidered too steep for sheet or wood pavement. It is not so durable 
as wood pavement, but does not require the constant repair of 
sheet-asphalt pavement. 

Brick pavement is inferior to the pavements above mentioned 
only because the noise of travel over the pavement is very much 
greater. It is unaffected by moisture or dirt, and, although it can 
be easily cleaned, does not need to be kept clean. It is therefore 
an ideal pavement for a small town or village where only a few 
streets are improved. It is less slippery in cold or wet weather 
than the asphalt or wood pavements. 

Stone-block pavements are rough, noisy, and hard to keep 
clean. For these reasons very few of these pavements are being 
laid. As the stones are very durable, and this was the only form 
of durable pavement in the country before 1875, it is probable 
that it will be long before they are wholly replaced. 

For further information consult Baker’s “Roads and Pavements”; 
Byrne’s “Highway Construction”; Herschel’s “Science of Road-making”; 
Aitken’s “ Road-making and Maintenance.” 


CHAPTER XXVI. 


WATER-SUPPLY. 

A complete system of water-works, designed to supply a 
town or city with water for domestic use, fire protection, and 
manufacturing purposes, consists of a collecting , a purifying , 
and a distributing system. 

The collecting system comprises the intakes where the water 
is taken from the source of supply, the receiving-reservoirs in which 
it is stored, the conduit through which it flows, and the pumping- 
machinery which may be necessary to raise it from one level 
to another. 

The purifying system comprises the settling-basins in which 
it is clarified, and the filters in which it is rendered pure and fit 
for use. 

The distributing system comprises the distributing-reservoirs 
in which the filtered water is received, and the network of pipes 
by means of which it is conveyed to the points of consumption. 

As will be shown later, all these elements are not found in 
every system of water-supply. 

A system of water-supply is either a gravity or a pumping 
system. 

A simple gravity system is one in which the intake is the most 
elevated point in the system, and no other force than that of gravity 
is required to move the water to and through the distributing sys¬ 
tem. The elevation of the intake should if possible be such as 
to produce a pressure at every point of the distributing system 
of about 45 to 120 pounds per square inch. If the pressure is 
too small, the water will not be raised to the upper floors of the 
buildings; if too great, it will cause leakage in the fixtures. The 
limits above given may be passed to meet local conditions. 

Pumping-machinery is introduced whenever the intake is not 

519 


520 


CIVIL ENGINEERING . 


high enough to produce the required pressure in the pipes, or 
when it is necessary to convey water over elevations over which 
it cannot be siphoned. The pumping-machinery may be located 
at the intake, or at a reservoir supplied by gravity from the intake. 
A pumping system is direct when the pumps force the water 
directly into the distributing system; it is indirect when they force 
the water into reservoirs from which it flows by gravity into the 
distributing system. 

The gravity system is the most economical to operate and 
the least liable to be interrupted by accidents. The direct pump¬ 
ing system is the most expensive to operate and the most liable 
to be interrupted by breakages. The indirect pumping system 
is more economical than the direct, especially for small plants, 
since the reservoir may be filled by day and the pumps stopped 
at night; it is also less liable to be interrupted, since the distributing 
system can draw on the reservoir whenever the pumps need repair. 

Collecting System. 

Character of Supply.—The ultimate source of every supply 
is the rain. The water which falls on the ground either flows 
along the surface until it reaches the ocean, or it sinks into the 
soil to form subterranean streams or reservoirs of ground-water, 
which usually have their outlets in surface-waters at a lower level. 
Much of the surface-water is lost on the way to its final destination 
by evaporation, or by sinking into the soil and becoming ground- 
water. Much of the ground-water is absorbed by vegetation. 

The immediate sources of supply are the ground-water of 
springs and wells, or the ground- and surface-waters mixed , of 
rivers , lakes , streams , and ponds. 

Springs are the natural outlets of an inclined porous satu¬ 
rated stratum of soil which rests on an impervious stratum or is 
enclosed between two such strata. Springs are found on hillsides 
where a porous stratum terminates, and at points where the upper 
impervious layer is broken by a geological fault and the head is 
sufficient to force the water to the surface. 

An artesian well is a spring formed by forcing a tube through 
an upper impervious stratum into a saturated porous stratum 
and thus making a path for the water to reach the surface. 


WATER-SUPPL Y. 


5 2 1 


A deep well is an artesian well in which the head is not suffi¬ 
cient to force the water to the surface when the upper impervious 
layer is pierced. If no very hard material is encountered, artesian 
and deep wells may be sunk by the water-jet process described 
under pile-driving. The water is forced to the bottom of the tube 
through an interior pipe, and not only softens the material there, 
but also washes out the earth between the two pipes. Through 
rock the hole is bored with a diamond or churn drill. 

A shallow well is a pocket made in a porous surface layer to 
collect the water for removal by pumps or other machinery. 
This well is usually either a large masonry shaft, like the common 
well, or a small metal tube sunk below the surface of the ground- 
water. To allow the water to enter freely, the bottom of the 
masonry shaft is open; the metal tube, which is 2 to 12 inches 
in diameter, is terminated by a perforated section or strainer 
which allows the water to collect in the tube. The tube is 
usually sunk by fitting its base with a conical cap and then driv¬ 
ing it like an ordinary pile; the well is therefore called a driven 
well. 

An infiltration-gallery is a long chamber of masonry or crib- 
work which is constructed across the line of flow of a water-bearing 
stratum which lies a short distance below the surface. It inter¬ 
cepts the flow of a large area of cross-section. It may also be 
constructed underneath the bed or in the banks of a running 
stream. 

Amount Required. —The amount of water consumed by 
any community depends primarily upon its population, and 
secondly upon the care which is taken to prevent waste. The 
amount used by factories is usually a small fraction of the total 
consumption. 

In England, where great care is taken to prevent waste, the 
average daily consumption is 33 gallons per capita. The average 
consumption in the principal cities varies from 21 to 61 gallons 
per capita. In this country, in cities where meters have been 
introduced to measure the amount supplied to each building, the 
average daily consumption is about 65 gallons per capita; this 
varies in the different cities from 30 to 100 gallons. If no attempt 
is made to control the wastage, the demand will usually be limited 
only by the capacity of the entire system. In many cities the 


5 22 


CIVIL ENGINEERING. 


daily consumption exceeds 200 gallons per capita, and in a few 
small communities the consumption exceeds even 300 gallons 
per capita. In this country, however, unless the conditions are 
exceptional, an allowance of 100 gallons per capita daily should 
be sufficient for any community in which the water furnished to 
each consumer is measured. 

The daily, weekly, and monthly consumption of water are 
by no means uniform. In warm weather much water is consumed 
in watering gardens and lawns; in cold weather much is allowed 
to run to waste to prevent the freezing of the pipes in the buildings. 
In this latitude, due to these causes, the maximum monthly con¬ 
sumption may exceed the mean monthly consumption for the 
year by 33 per cent; the maximum weekly consumption may 
exceed the mean weekly for the year by 50 per cent; the maxi¬ 
mum daily consumption may exceed the mean daily consumption 
for the year by 100 per cent. If, therefore, 100 gallons is con¬ 
sidered a sufficient average daily allowance per capita throughout 
the year, provision must be made in the system to allow this to 
be increased to 133 gallons in a single month, to 150 gallons in 
a single week, and to 200 gallons in a single day. The hourly 
consumption is also variable; the consumption for a single hour 
is sometimes 40 per cent more than the average hourly con¬ 
sumption for the day. 

The variations from the mean consumption are very much 
greater in small systems than in large ones; and larger in unmetered 
systems than in metered ones. The volume required by a steam 
fire-engine during a conflagration depends upon the character of 
the engine; it usually varies between 400 and 1200 gallons per 
minute; the amount thus drawn may be a heavy tax on a small 
system. 

Measuring the Supply.—If the source of supply is a spring, 
well, or small stream, the volume of the supply can be ascertained 
at any time by noting the time required to fill a reservoir of 
known capacity; or by noting the time required to restore, in a 
pool, a volume which has been removed by buckets of known 
capacity. 

The volume of discharge of a stream too large to be meas¬ 
ured in this manner may be gauged by causing the discharge 
to take place through an orifice or over a weir. 


WATER-SUPPLY . 


5 2 3 


A large river is gauged by measuring its mean velocity and 
its area of cross-section. 

Variation in Flow of Streams.—Because of the irregularity 
of the rainfall, the discharge or run-off of any drainage-basin 
or catchment area is a variable and not a constant quantity. In 
order that there may be no water-famine even in the period of 
extreme drought, the system of supply must be based on the 
minimum discharge of the stream which drains the basin. 

The most reliable method of determining the minimum dis¬ 
charge is to measure the daily discharge for a long series of years. 
The approximate methods which are ordinarily resorted to are 
either to estimate the discharge from the measured discharge of 
another basin subject to similar climatic conditions, or to esti¬ 
mate the discharge from the rainfall, employing the ratio of 
discharge to rainfall determined in some basin having similar 
climatic conditions. 

The Sudbury watershed near Boston, on which careful ob¬ 
servations have been made since 1875, is the standard employed 
for estimating water-supplies from watersheds east of the Mis¬ 
sissippi River. This basin has an area of about 75 square miles, 
of which 6 b per cent is water surface; the basin is generally 
hilly. The evaporation from the water surfaces in the basin is 
about equal to the rainfall. 

From the records of this basin it appears that— 

1. The run-off from the basin which may be utilized for 
w r ater-supply is about one-half the measured rainfall; or ap¬ 
proximately 1,000,000 gallons per day from each square mile 
of the basin. 

2. In a year of extreme drought the rainfall is two-thirds of 
its mean, and the discharge or run-off one-half of its mean. 

3. For five years in succession the rainfall and run-off may 
be less than the mean; for three successive years the rainfall 
may average only 80 per cent of its mean, and the run-off 70 per 
cent of its mean. 

4. The mean monthly rainfall is fairly uniform in this basin, 
being a minimum of 2.98 inches in June, and a maximum of 
4.57 inches in March. The mean annual rainfall is 45.83 inches. 

5. The average monthly run-off, on the contrary, is variable. 
It is equivalent to a depth of 5.17 inches over the basin in March, 


5 2 4 


CIVIL ENGINEERING . 


and only 0.35 inch in July. The run-off during February, 
March, and April is 50 per cent of the total annual run-off, and 
that from November 1st to May 31st is 85 per cent of the annual 
run-off. 

6. The minimum rainfall and the minimum run-off in a 
month may be only 10 per cent of their mean values. 

From observations on other watersheds it appears that the 
ratio of the run-off to the rainfall decreases as the amount of 
annual rainfall decreases. Hence in a basin in which the rain¬ 
fall is less than in the Sudbury basin the run-off will be less than 
one-half the measured rainfall. 

The difference between the run-off and the rainfall is due 
principally to evaporation and absorption by vegetation; water 
may also be lost by escaping from the watershed by subterranean 
channels. 

The annual evaporation from water surfaces depends on 
the local hygrometric conditions. In the United States the 
annual evaporation for places having an annual rainfall be¬ 
tween 30 and 50 inches may be either somewhat greater or some¬ 
what less than the rainfall, depending on local conditions; if 
the rainfall is very great, the evaporation is relatively small, and 
vice versa. At Astoria, on the Pacific coast, the rainfall is 77 
inches and the evaporation is 25 inches; at Cheyenne, Wyo¬ 
ming, the rainfall is 13 inches and the evaporation 76 inches. 

In calculating the run-off of a watershed where the rainfall 
and the evaporation are equal to each other, the area of all ponds, 
lakes, streams, etc., must be subtracted from the total area. 

The amount of water absorbed by the vegetation will vary 
from 10 to 15 inches if fed by the rainfall; some irrigated fields 
can absorb at least ten times as much. 

Variation in Flow of Springs and Wells.—The irregularity of 
the rainfall, especially its inequality over long periods of time, as 
seasons and years, must affect the underground streams and res¬ 
ervoirs in the same way that it affects the run-off of watersheds. 
The underground storage will be a maximum when the greater 
part of the annual rainfall occurs in the months when the soil is 
porous and the loss by evaporation is least. A mild wet winter is 
favorable to the undergound flow and storage of water, and a dry 
cold winter is unfavorable to it. 


WA TER-SUPPL Y. 


5 2 5 

The flow of a spring or well is not alone dependent on the 
amount of rainfall or the drainage area from which it receives 
its supply; it is also dependent on the extent of the underground 
area drained by the spring or well, and by the velocity of flow 
toward the outlet. 

Any difference of level between two points of a continuous 
water surface, whether above or below the surface of the ground, 
will cause a flow of water from the higher to the lower point; 
the velocity of the flow will vary directly with the difference of 
level between the two points, and will vary inversely with the resist¬ 
ance to the flow. 

The flow of a spring or well may be increased by increasing 
the depth of its water-level below the general surface of the 
ground-water. This will increase both the area drained and the 
velocity of flow towards the outlet. In an open porous water¬ 
bearing stratum the effect will be greater than in a compact 
one, since the resistance to the flow of the water through the soil 
is less. As a slight difference of head will increase the flow, the 
supply from springs and wells, in which the water-level may be 
lowered by pumping, is practically constant except in seasons of 
extreme drought. 

Receiving-reservoirs.—A reservoir is an artificial basin in 
which water is temporarily stored. Those belonging to the 
collecting system are called receiving-reservoirs; those to the 
distributing system are called distributing-reservoirs. All are 
also called storage-reservoirs. 

In a gravity system the purpose of a receiving-reservoir is to 
store the water flowing from an irregular source of supply and 
secure uniform daily flow into the distributing system greater than 
the minimum daily flow from the source of supply. In a pumping 
svstem it may also be employed to store the water which is pumped 
by an intermittent-acting pumping-plant, and secure a uniform 
flow into the distributing system from the receiving-reservoir. 

A receiving-reservoir is unnecessary only when the source of 
supply is a great river or lake whose minimum daily capacity is 
much greater than the maximum daily consumption; or when 
the source of supply is practically constant, as in the case of wells 
and springs. 

If the source of supply is the flow from a drainage-basin in any 


5 2 <3 


CIVIL ENGINEERING. 


of the States bordering the Atlantic Ocean, the average daily 
supply which can be collected from a square mile of area is about 
1,000,000 gallons. In a year of extreme drought, however, the 
average daily yield may be reduced one-half, or be only 500,000 
gallons. It is evident that if the distributing system requires more 
than 500,000 gallons daily from each square mile of the basin, 
there will be a water-famine during this year of drought, unless 
sufficient water has been stored in reservoirs to provide for this 
deficiency. It is also apparent that the nearer the daily con¬ 
sumption is to 1,000,000 gallons the greater must be the volume 
of water stored. If this year of extreme drought is preceded by 
one or more years in which the average daily yield is also less than 
the average daily consumption, an additional supply will be needed 
for these years. 

As it is not desirable to have reservoirs partially empty for 
too long a period, and as several years of drought may follow 
each other, it is usual to limit the maximum daily consumption 
to about 60 to 75 per cent of the average daily capacity, or to 
600,000 to 750,000 gallons daily per square mile of the drainage- 
basin. Reservoirs having a capacity of 200 to 250 days’ supply 
will then be sufficient to supply the deficiency of three successive 
years of drought. Even under favorable conditions it is difficult 
to secure sites for reservoirs to store greater volumes. 

If the reservoirs are made by damming one or more of the 
streams of a basin, as is usually the case when reservoirs of large 
capacity are required, they are also called itnpounding-reservoirs. 

If the average daily consumption is less than the average daily 
supply during the year of extreme drought, it will still be necessary 
to have receiving-reservoirs, if the maximum daily consumption 
is more than the minimum daily flow which has been recorded 
or may be expected in any summer or autumn month. From 
the Sudbury basin the average daily flow in July is only about 
one-fifth of the average daily flow for the year, or 200,000 gallons 
daily per square mile. In a year of extreme drought this may 
be reduced to one-tenth of this amount, or 20,000 gallons daily. 
If, therefore, the daily consumption exceeds 20,000 gallons per 
square mile of drainage area, some provision must be made for 
storage. To provide for an average daily consumption approach¬ 
ing 500,000 gallons, the total excess yield of the winter and spring 


WATER-SUPPLY . 


5 2 7 


months must be stored for use in the summer and autumn. The 
maximum storage required to equalize the flow for a single year 
is about 75 days’ supply. 

If pumping-machinery is employed to raise the water from 
a constant supply to a receiving-reservoir, the latter need hold 
only a few days’ supply to allow the engines to work intermittently, 
and as a safeguard against the temporary stoppage of the pumps 
due to accidents and breakages. Reservoirs of this type are made 
on some convenient elevation near the pumping-plant or purify¬ 
ing system. 

Dams.—The dams for impounding-reservoirs are made of 
earth or masonry. 

An earthen dam is usually of the type shown in Fig. 168. 



In the construction of an earthen dam care must be taken to 
prevent leakage through the dam itself, to prevent the water of 
the reservoir from rising to the height of its crest, and to prevent 
leakage underneath the dam or around its ends. 

Leakage through the dam is prevented by making a wide 
embankment with gentle slopes of carefully selected, thoroughly 
compacted material; or by constructing a thin impervious core- 
wall of puddle or masonry to resist leakage, and supporting this 
wall by an embankment of less carefully selected material (Fig. 
168). 

The water is prevented from rising to the level of the crest 
of the dam by constructing a spillway of sufficient capacity to 
discharge the water which flows into the reservoir at the time 
of maximum flood. 

Leakage under the dam and around its ends is prevented by 
removing all porous material and making a firm bond between 
the non-porous material of the site and that of the dam. 

A dam without a core is from io to 30 feet wide on top, 
depending on its height. The inner slope is 1/2, or if the mate¬ 
rial is soft, 1/3; this slope is usually paved or riprapped with 
























5 2 8 


CIVIL ENGINEERING. 


stone. The outer slope is 1/2, and is sodded to prevent its 
destruction by the rains. Both slopes, if long, are broken by 
berms. The material of the dam is of the same character as that 
employed in the coffer-dam, and is laid in thin layers which are 
moistened and then compressed by driving animals or moving 
grooved or studded rollers over them. 

A masonry core is a concrete or stone wall about 2 to 4 feet 
wide at the water surface, with a batter of 24/1; it is firmly 
imbedded in the impervious stratum upon which the dam rests. 
It must be water-tight. A puddle core is a similar wall of puddle, 
but is made twice as thick. The puddle wall is inferior to the 
masonry one in that it is more liable to crack and is not proof 
against burrowing animals. For these reasons the layers of 
the embankment immediately in contact with the faces of the 
puddle core are usually of selected material. In a dam with a 
core the main function of the embankment is simply to support 
the core; it may therefore be of less carefully selected material 
than in the dam without a core. When random rock is em¬ 
ployed to support the core the dam is called a rock-fill darn. 

Cores have been made of steel plate supported by a frame¬ 
work, and they have been replaced by an impervious apron of tim¬ 
ber, concrete, or metal placed over the inner slope of the dam. 

To prevent the leakage which would be liable to occur at the 
discharge-pipe which passes through the dam, especially if it 
settles, this pipe is laid on a bed of concrete resting on the im¬ 
pervious stratum beneath the dam. The pipe itself is then 
imbedded in concrete, and this covering of concrete has pro¬ 
jecting rings to increase the resistance to leakage along its sur¬ 
face. The pipe may also be laid in a small culvert which is 
constructed on the impervious layer, passes underneath the dam, 
and terminates in a vertical masonry tower, called a valve-tower 
or gate-house. A pipe thus laid is subject to constant inspection. 
(Fig. 168.) 

Spillway.—One of the most important features of an impound- 
ing-reservoir having an earthen dam is the spillway or waste- 
weir through which the surplus water is discharged when the 
reservoir is full. If this outlet is too small, in time of flood the 
water will rise above the crest of the dam, and flowing over it 
will soon cause its total destruction. If placed in the dam it- 


WATER-SUPPLY . 


5 2 9 


self and not properly constructed, the spillway itself may be 
washed out and thus destroy the dam. 

The length of the crest of the spillway, determined by the 
formula on page 302, should be such that it can discharge the 
maximum flood without causing the water in the reservoir to 
rise to a plane above the top of the core-wall or 3 feet below 
the crest of the dam. 

Various empirical formulas have been devised for the maxi¬ 
mum flood discharge; the most common is Fanning’s formula, 

D = 200.4^, 

in which T> = the discharge in cubic feet per second, 

A = the area of the basin in square miles. 

In this formula no factors are introduced which depend on 
the shape, slope, or character of the surface of the basin, although 
all these must affect the maximum discharge. It is thought to 
give too small a discharge for rapidly discharging basins in which 
the area is less than 10 square miles. The Burkli-Ziegler formula 
previously given is also used. 

The spillway should if possible be constructed in the solid 
ground beyond one end of the dam, and the water carried in a 
separate channel so far away from the dam that the backwater 
will not come in contact with the earth embankment. Unless 
this channel passes over a ledge of rock it should be paved to 
prevent scouring. If the spillway must be made in the dam it¬ 
self, the part of the dam containing the spillway should be of 
masonry; the channel for the water should be so inclosed that 
the water cannot reach the earthen dam. Spillways, in unimpor¬ 
tant dams, are made of cribwork filled with stones and covered 
by a plank sluiceway through which the water flows. The 
length of the weir of the spillway may also be determined less 
accurately by the Gould formula 

/= 20 ^ Ay 

in which l — length in feet, 

A =area of basin in square miles. 

Masonry and Rock-fill Dams. —A rock-fill dam is one whose 
stability depends on a wall of random rock. The faces of the 


53° 


CIVIL ENGINEERING. 


wall are given the natural slope of random rock dumped in place. 
Leakage may be prevented by covering the upper slope with an 
apron of well-calked planks or a sheet of steel plate, by a facing 
of concrete or masonry, by a thick layer of earth either above or 
below the rock dam, or by a core of masonry, reinforced con¬ 
crete, or steel plate. Leakage under the dam is usually prevented 
by the construction of a wall connecting the apron or core with 
the bed-rock. This type is employed in the construction of 
irrigation-reservoirs in our own country. 

Very high impounding-reservoir walls are usually made of 
masonry, resting on a natural rock surface. The principles 
governing their construction have already been discussed. The 
entire wall or a portion thereof may be used as a spillway. If the 
water wastes over the dam itself, the dam is usually made of the 
form shown in Fig. 169; this is called an ogee-jaced dam. The 

water follows the contour of the face and is dis¬ 
charged at the base parallel to the river-bed. 
This reduces to a minimum the tendency to 
erode the bed. 

Valve-tower, Gate-house, and By-pass.— 

A valve-tower , such as shown in Fig. 168, 
allows the water to be drawn from different 
levels of the reservoir; the water can be screened, however, 
only at the inlets. The screens or sieves cannot therefore be 
readily examined. 

A gate-house , whose exterior is similar to a tower, is usually 
divided into two vertical chambers. In the separating partition 
is an opening in which are placed a pair of movable vertical 
screens. The water enters the up-stream chamber through 
sluice-gates in its walls at different elevations, and after passing 
through the screens escapes through the discharge-pipe. Either 
screen can be removed at any time for cleansing. 

In both tower and gate-house there is usually a separate waste- 
pipe near the bottom for emptying the reservoir. 

The by-pass is a conduit connecting the inlet-pipe with the 
discharge-pipe so that the water may pass around the reservoir 
when it is emptied for any purpose. 

Intake.—If the source of supply is a stream, a dam is usually 
constructed at the intake so that the level of the water will be 






WATER-SUPPLY. 


531 


constant, and the inlet may be placed well below the level of the ice; 
the end of the conduit is covered by a suitable netting to prevent 
ingress of materials which might obstruct the flow through it. 
If the pool at the intake is a large one so as to be only slightly 
affected by the suction of the conduit, there will be little danger of 
the obstruction of the conduit by the needles of ice called anchor- 
ice. 

In the great lakes on our Canadian boundary the intakes are 
located in strong cribwork shafts which are far enough from the 
shore, if possible, to secure water unaffected by the city sewage. 

Conduits.—The main conduits of a system of water-supply may 
connect the intake directly with the purifying or the distributing 
system, or the impounding-reservoirs with either of the above 
systems. These may be constructed of masonry, steel, cast iron, 
or wood. Large conduits not exposed to internal pressure are 
usually made of masonry; they are simply covered canals, and are 
constructed with a .uniform grade of slight inclination. The 
Croton Aqueduct of New York, which connects the impounding- 
reservoir at Croton with the receiving-reservoir in New York 
City, is a masonry conduit whose cross-section is either a circle 
whose diameter is i2\ to 14 feet, or an equivalent horseshoe sec¬ 
tion; the uniform slope of the larger section is seven-tenths of a 
foot per mile. Large conduits which are exposed to considerable 
internal pressure are made in the form of riveted steel pipes; 
conduits from 4 to 6 feet in diameter have been thus constructed. 
Cast-iron pipes are more commonly employed than any other form, 
and are manufactured in sizes varying from 4 inches to 5 feet in 
diameter. Conduits made of wooden staves with steel bands are 
often constructed where suitable lumber is cheap and the pressure 
in the conduit is not excessive; conduits 6 feet in diameter have 
been made of wood. 

Pumping-machinery.— Pumps are of a great variety of forms, 
depending upon the volume of water to be raised in a unit of time, 
the head or back pressure against which they must work, and the 
depth of the water-level below the surface of the ground. The 
back pressure is constant if the water is pumped into a reservoir, 
and variable if pumped directly into the distributing-pipes. One 
or more reserve pumps are usually installed to provide for emer¬ 
gencies; these are absolutely essential in a direct pumping system. 


532 


CIVIL ENGINEERING. 


Water is sometimes lifted from wells to the surface by forcing 
compressed air to the bottom of the vertical collecting-pipe through 
a small interior pipe. The mixed air and water rise to the surface 
in the outer pipe. 


Purifying System. 

The object of the purifying system is to remove the inor¬ 
ganic material in suspension which makes the water turbid, the 
mineral matter in solution which gives it undesirable qualities, 
the organic substances in suspension which are the food of patho¬ 
genic bacteria or disease organisms, these organisms themselves, 
and finally the organic matter which colors the water. 

The first is principally the mud found in streams derived 
from surface drainage; the turbidity of such streams is exceed¬ 
ingly variable, being very great after severe rain-storms and very 
slight in times of drought. Mud may be removed by allowing 
the water to remain quiescent in settling-basins for a long period 
without the assistance of coagulants, or to remain a short time 
with coagulants. The process of clarifying may also be partially 
made in the settling-basins, and completed in the filters. 

The inorganic properties in water, such as lime and iron, which 
give it undesirable qualities, are removed by special processes 
as described hereafter. 

The organic substances which furnish food for pathogenic 
bacteria and the bacteria themselves are to a great extent removed 
in a settling-reservoir if the water is allowed to remain in it long 
enough, perhaps for a month or more; as a rule, however, to 
remove bacteria the process of sedimentation is followed by that 
of filtration. As it is impossible at present to destroy the patho¬ 
genic bacteria without at the same time destroying the harmless 
varieties, the purity of the water is measured by the total number 
of bacteria in it If the number of bacteria in a cubic centimeter 
does not exceed one hundred, the water, while not absolutely 
pure, is considered healthful within reasonable limits. This 
result is obtained in ordinary river-water only by the destruction 
of over 99 per cent of the bacteria in it. 

The organic coloring-matter is usually removed in the pro¬ 
cesses of sedimentation and filtration. 


WATER-SUPPLY. 


533 


Selection of Sources of Supply. —As it is impossible to remove 
all the bacteria from water, a source of supply should be selected 
which is not liable to contain pathogenic bacteria. Waters have 
been classified as follows: 

\ 

1. Water from springs, deep wells, and the surface drainage 
of uncultivated and unpopulated lands are classed as whole¬ 
some. 

2. The surface drainage of cultivated and sparsely populated 
land, and rain-water drained from roofs, are classed as suspicions. 

3. Waters from rivers into which sewers empty, from shallow 
wells, and from fields enriched by night-soil are classed as dan¬ 
gerous. 

Ground-waters, except those of the last class, are usually 
wholesome as far as bacteria are concerned, but are liable to 
contain inorganic substances which may be objectionable or 
even dangerous. Ground-waters are also more liable than sur¬ 
face-waters to nourish vegetable growths wdiich give the water 
an offensive odor or an objectionable taste. These may be 
prevented by keeping the water under cover from the time it 
reaches the surface of the ground until it passes into the dis¬ 
tributing mains. 

Settling-basins. —The term settling-basin is ordinarily applied 
to a reservoir which contains from one to four days’ supply of 
water; in such a reservoir the water derived from surface drainage 
is made clear enough to prevent it from clogging the filter-beds. 
The bed of the basin is of concrete, so that it can be thoroughly 
cleaned when necessary. The depth of the water in the basin 
is usually from 8 to 15 feet. A less depth does not prevent the 
growth of vegetation, and a greater depth makes the time required 
for sedimentation too great. Sedimentation covering such a short 
period has no important effect in reducing the number of bac¬ 
teria. 

The flow of the water through the basins may be either con¬ 
tinuous or intermittent. In a continuous system, if there is but 
a single small basin, the water enters near the bottom of one end, 
moves across the basin with a very slight velocity, and leaves the 
basin near the top of the opposite end. The reservoir can be 
cleaned only by conducting the water through a by-pass or pipe 
which connects the conduit above with the conduit below. The 


534 


CIVIL ENGINEERING. 


water may be compelled to pursue a serpentine course through 
the reservoir by inserting vertical plank partitions called 
baffles. 

To clean the basin a day is selected upon which the water 
received from the source of supply is unusually clear. The 
inlet into the reservoir is closed, and the water is allowed to flow 
around it through the by-pass. The basin is then emptied by 
means of an outlet constructed for this purpose. If there are 
two or more basins, the water flows through them all except when 
it is desired to isolate and empty one of them. The water then 
flows around it through its by-pass. 

In an intermittent system the basins are arranged in series of 
three. One of the three is the receiving-basin. Another is the 
settling-basin, in which the water remains quiescent while the 
first is being filled. The third is the basin from which the settled 
water flows into the distributing system. 

If there is a large receiving-reservoir in the system, it will 
perform all the functions of a settling-basin better than a special 
basin, as the water remains in it a longer time and its depth is 
usually greater than that of a basin. Such a reservoir has also 
the effect of materially reducing the number of bacteria. 

The turbidity of water is tested by noting the distance below 
the surface at which some standard object can be seen. This 
object is usually a fine platinum wire fastened at right angles to 
the axis of a rod graduated to millimeters. 

Filters.—The object of filtration is to complete the purifica¬ 
tion of the water begun in the settling-basin or the receiving- 
reservoir. Filters are often divided into two classes; one is the 
slow, sand, or English filter, and the other is the rapid, mechanical , 
or American filter. 

Slow Filters.—The ordinary slow filter is a basin into which 
the water flows through an inlet above the filter-bed, and from 
which it escapes through an outlet below the filter-bed. The 
basin may be open or covered. In a cold climate it is usually 
considered advisable, though not absolutely necessary, to cover 
the basin and thus prevent the water from freezing; the removal 
of the ice is expensive and the filter works less thoroughly when the 
water is very cold. In a warm climate some authorities prefer 
a covered basin because there is less vegetable growth in it, while 


WATER-SUPPLY . 


535 


others prefer the action of the sun-rays on the water because of its 
bacteria-destroying effect. 

The basin is made in two or more divisions, so that a single 
division may be emptied and cleaned without interfering with the 
filtration process. 

The slow filter as designed by Mr. Hazen, C.E., for Albany, 
Washington, and other places, consists of a basin with a bed 
and vertical side walls of concrete. The roof consists of groined 
arches resting on pillars of brick or concrete. The pillars are 
about 9 feet high and 20 inches square, and are placed in rows 
which are 14 feet apart in each direction. A section through two 



pillars is shown in Fig. 170. By substituting a continuous wall 
for a row of pillars, the basin is divided into two or more separate 
parts. The roof of each groined arch is pierced by a manhole 
for light and for conveying material; the concrete roof is covered 
with earth. 

The filter-bed proper consists of a bed of sand about 3J feet 
thick. The specifications require that the sand shall be fine; 
no particle must be over 5 mm. in diameter, and 70 per cent must 
not exceed 1 mm. in diameter. Around the collecting-pipe is a bed 
of gravel varying in size; the finest gravel is near the sand. The 
gravel prevents the sand from being washed away. 

Each division of the filter has a large collecting-drain which 
is laid below the level of its floor and bisects its area. Upon 
the floor of the filter and at right angles to the main drain are 
the branch drains which collect the filtered water. The main 
and branch drains are stoneware pipes; the latter are laid with 
slightly open joints. From the main drain the water passes by 











































536 


CIVIL ENGINEERING. 


gravity or is pumped to a covered clcar-water basin. Fig. 171 
shows one chamber of a filter, in which the heavy black line is 
the main drain and the lighter cross lines are the tributary drains. 
The roof is covered by a layer of earth about two feet thick. 
The top of the covering is usually sodded. 

To prevent the disturbance of the bed of sand by the force 
of the current from the inlet, the inlet is in a small chamber 
separated from the main floor-area by a low wall over which 
the water flows; the wall checks its velocity. To prevent dis¬ 
turbance when the filter is started after being cleaned, provision 









□ 


□ □ 

□ □ 


□ 


0 □ 

□ □ 


□ 


□ □ 

□ □ 


□ 


□ 

□ 

□ 

□ 










Fig. i7i. 


is made for allowing filtered water to enter from the drains. 
Regulators are also provided for controlling at all times the head 
or depth of water over the filter-bed. 

The filter is started by admitting water from below until the 
bed is flooded; the water is then allowed to enter through the 
regular inlet. The head is regulated to make the rate of filtra¬ 
tion such as will produce an effluent of the desired purity. The 
rate which is usually considered safe is a velocity of flow verti¬ 
cally through the filter-bed of 4 inches per hour, or 8 feet per day. 
This is equivalent to about 2,600,000 gallons per acre per day. 
Experiments have shown that this rate may be increased for clear 
and slightly polluted waters to 6,000,000 gallons per acre per 
day without producing an effluent which is bacterially impure. 

After a time the surface of the sand becomes covered by a 
gelatinous coat which reduces the velocity of flow and requires 
the head to be increased. When the limiting head has been 
reached, and the rate is still further reduced, the filter is emptied 
and scraped. In this operation the surface layer to a depth of 























WATER-SUPPLY. 


537 


an inch, more or less, is removed and the filter is again ready for 
work. The scrapings are carefully washed, and when the thick¬ 
ness of the sand is reduced to about 2 feet the washed sand is 
again placed in the filter. 

The process of the filter is both mechanical and chemical. 
The large organic and inorganic bodies are removed as by a 
sieve; the minute organic substances are converted into their 
elements, principally at the gelatinous coat, by the aid of bacteria; 
the bacteria disappear with the substances upon which they 
thrive. The gelatinous coat is therefore considered the most 
valuable feature of the filter, and care is taken not to rupture it 
by a too rapid change of velocity of flow. The bacteria engaged 
in the reduction are dependent on oxygen. If the supply of 
oxygen in the water is sufficient, the process of filtration may be 
continuous; if the supply is not sufficient, the process is inter¬ 
mittent, and the filter is emptied at regular intervals to allow the 
air to penetrate the pores of the bed. 

To prevent the rapid clogging of the sand-bed, the water is 
sometimes passed through a preliminary coarse filter or screen 
to remove the coarse particles. This is called a scrubber. 

Rapid Filters.—Filtration as above described is necessarily a 
slow process, and a large area is required for the filter-beds. To 
increase the rapidity of filtration, filters have been devised in 
which a coagulant is employed to remove the inorganic materials, 
and the sand is aerated by mechanical means. The coagulant, 
usually sulphate of alumina, forms a gelatinous coat over the 
sand, which performs a similar function to that in the slow filter. 
The water is forced through the filter by its natural head if suffi¬ 
ciently high, otherwise by pressure. When the flow is checked 
by the sediment on the bed, the sand is washed by reversing its 
current and at the same time stirring it by a current of air or by 
mechanical means. When thoroughly washed the process of 
filtration is repeated. In the rapid filter, therefore, the process 
is intermittent. The rate of filtration is about 125,000,000 gallons 
per acre per day, or many times that of the slow filter. Many of 
the mechanical filters are in the form of wooden or metal tanks 
having a filter-bed whose area is about 10 square feet. The 
sand is coarser than that employed in the slow filters and rests 
upon some form of sieve. If the water is very turbid, it is clarified 


53S 


CIN/L ENGINEERING. 


to some extent in a clarifying-tank before it is allowed to enter 
the filter proper. Care must be taken to regulate the supply 
of alum so that there shall be no free sulphuric acid in the water; 
sometimes lime is added to precipitate the acid. 

Iron may be removed from water by first oxidizing it by 
aeration and then passing it through a sand filter. This process 
is employed to purify water from wells. If the water does not 
contain organic matter, the process may be a rapid one. 

Temporary hardness in water is due to the presence of cal¬ 
cium carbonate in water which contains free carbonic acid. The 
carbonate is soluble only when the carbonic acid is also present, 
and may therefore be precipitated by any process that removes 
the acid. The carbonic acid may be removed from small quan¬ 
tities by boiling the water; from large quantities, by adding lime- 
water, with which the carbonic acid can unite and form additional 
carbonate and thus remove the free carbonic acid. The carbon¬ 
ate will then be deposited. Filters have been designed for ex¬ 
pediting this process, which is usually a very slow one. 

Distributing System. 

Distributing-reservoir.—A distributing-reservoir usually serves 
one or more of the following objects: 

1. It is a source of supply for the distributing system to pre¬ 
vent a water-famine in case of accident to any part of the system 
of supply between the intake and the distributing-reservoir. In a 
pumping system it is perhaps more necessary in this sense than 
in a gravity system. A reservoir of three or four days’ supply 
is usually considered of sufficient capacity for this purpose. 

2. It is a basin for the storage of the filtered water which 
flows from the purifying system. A reservoir of this type is also 
called a storage-reservoir. To reduce the probability of con¬ 
tamination of the water in a storage-reservoir to a minimum, 
the reservoir is usually limited to a day’s supply or less, and the 
reservoir is wholly inclosed and covered. 

3. It allows the hourly consumption to be variable, although 
the hourly supply which flows from the receiving-reservoir or 
from the purifying system is constant. As stated before, this 
hourly consumption may be nearly 40 per cent in excess of the 
mean hourly consumption. To furnish a reservoir for this pur- 


» 


WATER-SUPPLY. 


539 


pose the capacity of the reservoir is based on the hourly variation 
on the day of maximum daily consumption. In small plants 
the maximum hourly consumption is at the time of a large fire, 
and the reservoir is constructed to meet the emergency. A 
reservoir to meet the hourly variation in the consumption need 
never contain a maximum day’s consumption. 

Water-towers and Stand-pipes.—In pumping systems the 
distributing-reservoir may be a small tank placed on a skeleton 
steel tower, or it may be a vertical cylinder of riveted steel 
plate, called a stand-pipe. The object of the former is simply 
to regulate the back pressure, while the latter is in addition a 
small reservoir. Both the filling and discharging pipes enter 
the cylinder at its base. If the water-supply belongs to the class 
designated as wholesome and the receiving-reservoir is a large 
one, settling-basins and filters are often omitted and the receiving- 
reservoir serves also as a distributing one. 

Water-mains.—The system of distributing-mains should ful¬ 
fill the following requirements: 

1. Every part of the system should be strong enough to resist 
the static head of the distributing-reservoir. 

2. The hydraulic head at every point of the system should 
at all times be sufficient to raise the water to the desired height 
in the adjoining buildings. 

3. The water in every part of the system should be in con¬ 
stant motion, and should not be allowed to become stagnant. 

4. If a break or leak occurs in any main, it should be possible 
to stop the flow through a short section in the immediate vicinity 
of the break without interfering with the supply in other parts 
of the system. 

5. It should furnish an abundant supply to fire-plugs placed 
at intervals along the mains. 

These requirements are fulfilled in general by laying, in the 
streets of a city, a network of connecting-mains, thus forming an 
underground reservoir of considerable capacity, Fig. 172. The 
smallest of these mains has a diameter of at least 4 inches. The 
aggregate area of cross-section decreases with the decrease of 
the area to be supplied. 

If in the low part of the city the hydraulic head is too great, 
it is decreased by partially closing a valve in the main which 


540 


CIVIL ENGINEERING. 


supplies it, or a special valve may be introduced to cut off the sup¬ 
ply when the hydrostatic head reaches a certain amount. 

If in a high part of the city the head is too small, it may be 
increased within limits by supplying that section by a special 
main of large cross-section. If this does not give the desired 
head, the section is provided with a special pumping system. 



Fig. 172.* 


This is called the high service , in contradistinction to the lam 
or gravity service. 

By connecting the ends of the mains, the stagnant water which 
is always found in dead ends is avoided. By connecting the 
mains at intervals the head is made more uniform, and it is pos¬ 
sible to stop the flow in a short section of a main without mate¬ 
rially affecting the supply along the other main. It also in¬ 
creases the volume which may be drawn in a given time from any 
point in the system, as at a fire-plug. 

The water-mains of cities are usually cast-iron pipes with 
bell and spigot joints which are closed with hemp and lead. 
Valves are inserted at intervals so that any part of a main can be 
temporarily closed at both ends in order to make necessary re¬ 
pairs. The valves are vertical leaves which are lowered by means 
of a screw; this prevents their being closed rapidly. If the flow 
of water through a pipe is suddenly checked, the pipe is sub¬ 
jected to a severe blow from the water-hammer thus formed. The 


* A plan of a part of water-main system in Washington, D. C. 


































































































































WATER-SUPPLY. 


541 


intensity of the blow varies directly with the volume and velocity 
of the flowing water. 

Small house connections are generally lead or galvanized-iron 
pipes, usually about j inch in diameter, which can, by a special 
device, be attached to the main without stopping the flow of 
water. 

Fire-plugs are attached to short branches connected with the 
street-mains at the street-corners and in the interior of long 
blocks, and also for flushing purposes at the dead ends of the 
mains. These have studs to which the fire-hose can be 
attached. 

Blow-off.—A blow-off is a short branch with a suitable valve, 
through which the water in the main can be discharged. Blow- 
offs are placed at low points and at the dead ends of conduits 
and mains to force out deposits which may collect there. 

Air-escape Valves.—These valves are placed at the high 
points of conduits to remove the air which may collect there 
and interfere with the flow. 

Water-meters.—Meters of different patterns are employed 
for measuring the house consumption; they are attached to 
the service-pipes. The Venturi meter is one which may be em¬ 
ployed for measuring the flow under pressure through a main or 
conduit of any size. 

For further information see Merriman’s “Hydraulics,” Bovey’s “Hy¬ 
draulics,” Folwell’s “Water-supply Engineering,” Burton’s “Water-supply 
of Towns,” Turneaure and Russell’s “Public Water-supplies,” Goodelhs 
“ Water-works for Small Cities and Towns. 


CHAPTER XXVII. 


SEWERAGE. 

The term sewerage is applied to the removal of the liquid 
and soft solid waste products of a community, through a system 
of conduits, by means of its water-supply. The conduits are the 
sewers , and the waste material which flows through them is the 
sewage. 

The principal sources of the sewage are the household and 
factory wastes, and the surface drainage after rainfalls. If the 
household and factory wastes are carried in one system of con¬ 
duits and the surface drainage in another, the sewerage system 
is a separate one; if the wastes and drainage are carried in the 
same conduits, the system is a combined one. 

The principal advantages of the separate system are: 

1. Sewers for carrying the household and factory wastes may 
be constructed while the surface drainage is still carried in sur¬ 
face drains; the storm-water sewers may be added at a later 
period. This reduces the first cost of the system. 

2. If the sewage must be treated before it is finally discharged 
into streams and lakes, the separate system is the more economi¬ 
cal, since only the sewage of the house-sewers must be treated; 
that of the storm-sewers can be discharged without treatment. 
In the combined system the entire sewage must be treated until 
the dry-weather flow is sufficiently diluted by storm-water to allow 
it to be discharged without danger of creating a nuisance. 

3. In the separate system only the house-sewers need be deep 
enough to drain the basements of houses; the storm-sewers need 
only be deep enough to remove the surface drainage. In the 
combined system all the sewers must be deep enough to drain 
the basements. 

The principal advantages of the combined system are: 

542 



SEWERAGE. 


543 


1. Only a single system of conduits is in each street. This 
prevents confusion and possible errors in connecting houses with 
sewers. 

2. The periodical flushing of the sewers by the storms cleans 
them thoroughly. In the separate system reliance must be 
placed on automatic flushing-tanks, assisted by connecting small 
drainage areas, as roofs and yards, with the house-sewers. 

The separate system is usually better than the combined 
one for small communities whose surface drainage can be carried 
largely in open drains, and for all communities whose sewage must 
be treated before it can be discharged into the natural drainage 
streams of the country. It may also be advantageous when the 
sewage must be pumped to its final destination. 

The combined system is usually best for a large city, whose 
sewage may be discharged without treatment into the ocean or 
into a large river. 

Household Wastes.—The household wastes constitute the 
most important element of the sewage because of their offensive 
character and rapid decomposition. The aim of a sewerage 
system is: 

1. To remove in closed conduits from each house, and from 
the limits of the community itself, all household wastes before 
their decomposition begins. 

2. To seal every opening in the house-pipes and often, in ad¬ 
dition, the connection between the house-pipes and sewer, so that 
no gaseous products of decomposition generated in the pipes or 
sewer can enter the house. 

3. To make the sewers water-tight to prevent both the pol¬ 
lution of the surrounding ground by sewage and the infiltration 
of ground-water. 

4. To ventilate the sewers or otherwise remove offensive odors. 

5. To finally dispose of the sewage in a manner which will 
not produce a nuisance, or a menace to public health. 

House-drainage. 

The general system of house-drainage (Fig. 173) consists 
of one or more vertical soil- and waste-pipes which receive the 
wastes from the fixtures and which empty into the house-drain. 

The house-drain is usually an extra-heavy cast-iron pipe with 


544 


CIVIL ENGINEERING. 


leaded joints. For ordinary houses it is 5 inches in diameter 
and has a fall of at least 1 inch in 4 feet. If the basement is to 
be drained, the drain is laid on a bed of concrete, true to grade, 
below the floor of the basement. If the basement is not to be 
drained, the house-drain may be supported on masonry pillars, 
on brackets attached to the walls, or by stirrups attached to the 
floor-beams. Steel and wrought-iron drain-pipes are allowed 
when the drains are above the basement floor. The house-drain 
is provided with cleaning-holes at intervals, and empties into 
a stoneware pipe called the house-sewer, a short distance beyond 



the exterior walls of the house. A soil-pipe is a vertical pipe 
which receives the discharge of a water-closet. It is a heavy metal 
pipe, usually cast iron, which extends from the house-drain to a 
point about 2 feet above the roof. At its base it is supported 
on a masonry pier; its upper end either is left open or is closed 
merely by a wire cage to prevent the ingress of materials which 
might obstruct it. For ordinary houses the soil-pipe is 4 inches 
in diameter. In warm climates the soil-pipe is often placed on 
the outer face of one of the exterior walls of the building, to pre¬ 
vent air, escaping from a leak, from entering the house. 

A waste-pipe is a vertical pipe which receives wastes from 
other sources than water-closets. It is also a metal pipe supported 
below and terminated above like the soil-pipe. The diameter 
of a waste-pipe is usually 3 inches, but the length above the roof 
is enlarged to 4 inches in diameter to prevent its being closed by 
frost. 





























SEWERAGE. 


545 


The drainage of the roof is frequently conducted to the drain¬ 
pipe by a vertical iron pipe, called a leader , which is placed against 
the outer or the inner face of an exterior wall. For the ordinary 
house this is also 4 inches in diameter. 

« 

The pipes which carry the wastes from the fixtures to the 
soil- and waste-pipes are called branches. These are smaller 
than the verticals, and are usually made of lead, as lead pipe 
can be readily bent to the proper shape; branches are also made 
of nickel and brass. 

The general principles governing house-drainage are, to make 
the system of pipes simple and short , to avoid concealing the pipes , 
to make them readily accessible , and to provide for their ventilation 
and for the removal of obstructions. 

Traps.—A trap is any contrivance which, inserted in a pipe, 
will automatically prevent the passage of air or gas. The ordi¬ 
nary form of trap is a U-shaped- pipe 
filled with liquid (Fig. 174). The depth 
of the seal is the difference of level of 
A and B, which does not ordinarily ex¬ 
ceed 4 inches. It is evident that no 
air can pass the trap so long as the U- 
shaped pipe is filled above the level 
of B. 

This water-seal may be removed in 
one of the following ways: 

1. By evaporation. If the liquid in 
the trap is not renewed from time to 
time, it will gradually evaporate. This occurs in houses which 
are unoccupied for some time. To prevent this, the water may be 
replaced by some oily liquid like glycerin, which is not easily 
evaporated, or the inlet may be plugged if the fixture is not to 
be frequently flushed. 

2. By velocity. The velocity of the flush may be so great 
that the friction in the trap may not be sufficient to retain the 
water required for the seal when the flow is stopped. The nearer 
the trap to the inlet, the less is the probability of its being un¬ 
sealed in this way. 

3. By capillary attraction. If a piece of cloth or waste is 
caught in the water of the trap and the other end extends into 



Fig. 174. 






















546 


CIVIL ENGINEERING. 


the branch CD, the water will be slowly drawn into the branch 
CD by capillary attraction. 

4. By difference oj pressure. If the pressure due to a current 
of air is greater in CD than at A, the water will be forced up 
the tube AB towards the inlet, and the air will escape as soon 
as the lower surface of the water reaches B. 

5. By siphonage. If the pipe from C to D runs full, a siphon 
will be formed which will drain the trap. This is most liable 
to occur when the branch is of uniform diameter throughout 
and the length CD is great. It can be prevented by making an 
air-inlet above C as shown in the figure. 

The vent-pipe to which the air-inlet is connected is carried 
to a point above the roof, or is connected with a soil- or waste-pipe 

above the highest branch. As vent-pipes have 
the disadvantage of complicating the system, 
non-siphoning traps are now generally em¬ 
ployed. In these (Fig. 175) the trap is sud¬ 
denly enlarged or so obstructed as to check 
the velocity of flow through it. 

There are many forms of traps. Those in 
common use on waste branches are of lead, 
and if of the general form shown in Fig. 174 
are called S traps. The best are of drawn 

Fig. 175. lead, which makes them perfectly smooth inside; 
at the base is a cleaning-cap. 

A running or U-shaped trap is usually placed in the house- 
drain near its junction with the house-sewer (Fig. 173). This 
is omitted when it is desired to ventilate the sewer by means of 
the soil- and waste-pipes. A trap is also placed on every branch 
pipe as close as possible to its inlet. 

The house-drainage system is ventilated by carrying the soil- 
and waste-pipes above the roof and by making a fresh-air inlet 
into the house-drain in rear of the running trap near its outlet, 
as shown in Fig. 173. The draft through the vertical pipes may 
be increased by placing them near a chimney in constant use. 
The fresh-air inlet must not be placed near the cold-air shaft of 
a furnace, nor should this inlet or the upper end of a soil- or 
waste-pipe be near a window. 









SEWERAGE. 


547 


Sewers. 

The term sewer is usually applied to a common conduit which 
carries the drainage of several houses. However, in some city 
regulations the stoneware pipe which connects the house-drain 
with the common sewer is called the house-sewer, and in others 
the term sewer is applied to all conduits which carry household 
and factory wastes, in contradistinction to those which carry sur¬ 
face drainage only; the latter are then called drains. 

General Principles of Construction.—The sewer which re¬ 
moves the drainage of a block of houses differs from the water- 
main which supplies it in the following particulars: 

1. The water-main is always under sufficient head to raise the 
water to a considerable elevation above the street. The sewer should 
never be under a head sufficient to force the traps in the house-drains. 

2. From the above it follows that while the water-pipe may 
be laid in any convenient manner, providing no point is higher 
than the hydraulic grade-line, the sewer must be laid with great 
care so that the slope of its bed shall give the required velocity 
of flow without producing pressure on its crown. 

3. The depth of the water-main is immaterial so long as it 
is great enough to protect the service-pipes from frost. The 
sew T er, however, must ordinarily be low enough to drain the base¬ 
ment of every house, and also to receive the sewage from all 
tributary sewers which are nearer the head of the system. Its 
slope will usually be approximately that of the street, unless this 
slope is less or greater than the limits for sewer grades. 

4. If all the water-taps in the block are opened and allowed 
to discharge into the sewer, the velocity of flow in the water- 
main will be much greater than in the sewer, due to its greater 
head. Since the water-main which supplies and the sewer which 
discharges must provide for the same volume, the diameter of 
the sewer must be much greater than that of the water-main. 

5. While the w T ater-main carries a practically perfect liquid, 
the sewer always carries a liquid charged with solids which impede 
its flow. In sewers carrying surface drainage the sewage con¬ 
tains sand, gravel, sticks, etc. The sewer is therefore much 
more liable to be obstructed, and special provision must be made 
for the removal of obstructions. 


54§ 


CIVIL ENGINEERING. 


6. The character of the sewage demands also its removal to 
the place of final disposal in as short a time as possible, and the 
ventilation of the conduit in which it is carried. 

Material of Construction—The materials of which a sewer 
is constructed are subjected to external rather than internal 
pressure, and only need be strong enough to resist this pressure. 
The inner surface of a sewer should be smooth, so as to offer 
little resistance to flow, and it should not be acted upon by the 
acids of the sewage. There should be as few joints as possible 
in the sewer, and the material should lend itself to the construc¬ 
tion of curves at the points where the sewer changes direction, 
so as to reduce the resistance at these points to a minimum. 

The materials which best fulfill these conditions are stoneware 
pipes, glazed on the interior, concrete pipes, and brick conduits 
lined with some smooth surface. Iron pipe is employed if the 
sewer is under a head, as an inverted siphon sewer under the bed 
of a river which connects the sewers on the opposite banks; iron 
pipe is also sometimes employed in very compressible soil where 
the stoneware pipe would be liable to break. 

Glazed Stoneware or Vitrified Pipe.—The standard sizes of 
vitrified pipe vary from 2 to 24 inches in diameter; special pipes 
are made of larger diameters. Except for house-sewers, which 
may be 4 or 5 inches in diameter, the smallest pipe employed 
in sewer construction is the 6-inch pipe. In a combined system 
the smallest sewer is usually the 8-inch pipe. Stoneware pipes 
of a diameter exceeding 24 inches are not often employed. 



Fig. 176. 


Sewer-pipes are made of two general forms; one is called 
the socket , spigot-and-bell , or spigot-and-hub pipe , and the other 
is the ring pipe (Fig. 176). The latter is a simple cylinder, and 
is provided with rings or bands to cover the joints. A channel 
or split pipe is a semi-cylindrical pipe. Increasers and reducers 
are conical-shaped pipes employed to unite sewers of different 
radii; the former have the socket at the small end, and the latter 
at the large end. 





















SEWERAGE. 


549 


In making a joint in a socket-pipe, the spigot of one pipe is 
inserted in the socket of the other and surrounded by a gasket 
of waste. The socket is then filled with neat cement or mortar 
well compressed. Experiments indicate that neat portland 



Fig. 177. 


makes the tightest joint; one-to-one portland is next in order, 
and one-to-one natural cement is third. The waste prevents 
the mortar from escaping into the pipe and making a rough sur¬ 
face. The ring pipe is usually laid in a bed of concrete; the 
joints are covered with concrete or with bands, and the bands 
themselves are buried in concrete. Fig. 177 shows the method 
of laying ring sewers in Washington, D. C. Section A is at a 
joint and section C between joints. 

Brick and Concrete Sewers.—Sewers whose area of cross- 
section exceeds that of the 24-inch pipe are commonly made of 



brick or concrete. The usual forms of cross-section are the 
circular , the egg-shaped , and the horseshoe. The circle gives the 
greatest area of cross-section for the same amount of material 
and is therefore the cheapest; the egg-shape is employed to in¬ 
crease the velocity of flow in a combined sewer when the dry- 


































55° 


* CIVIL ENGINEERING. 


weather flow is small as compared with the storm flow. The 
hydraulic mean depth, and .hence the velocity, is greater in the 
egg-shaped sewer when only slightly filled than in the circular 
sewer of the same area of cross-section. The horseshoe-shaped 
sewer is employed when it is desired to secure a large area of 
cross-section without an excessive height. Fig. 178 shows a 
horseshoe-shaped sewer constructed in Washington, D. C. To 
reduce the amount of friction, the lower part of a brick sewer 
is often made of blocks of glazed vitrified stoneware, called sewer- 
ox invert-blocks. To reduce the wear on the beds of sewers with 
steep slopes, the beds are sometimes lined with vitrified bricks or 
stone paving-blocks. 

Ilanholes.—A manhole is a vertical masonry shaft which 
gives access to a sewer from the surface of the site, either for 

inspecting or for cleaning the sewer. 
The top of the shaft is usually 
circular and 2 feet in diameter; 
the bottom of the shaft is either 
circular, oval, or square, and is large 
enough to allow the sewer to be in¬ 
spected and cleaned with ease. For 
this purpose the larger dimension need 
not exceed 5 feet. It is desirable to 
have a manhole at every point where 
a sewer changes its direction or slope, 
so that every stretch can be readily 
inspected. In the straight stretches 
it is also desirable to have a man¬ 
hole every 100 yards, but this distance 
may be increased or diminished. 
The top of the manhole is usually 
closed with a perforated iron cover, 
and some form of ladder is attached 
to its walls to enable the workman to 
pass up and down. The shaft need 
not be directly over the sewer; some¬ 
times it is placed on one side and 
connected with the sewer by a hori- 



\t£i izMz&s 
Fig. 179. 


zontal passage. Fig. 179 shows a manhole on an egg-shaped 













































































































SEWERAGE. 


551 


sewer as constructed in Washington, D. C. The ladder is made of 
U-shaped iron bars set in the walls of the manhole (C, Fig. 179). 
The horizontal section at A is circular, and at B rectangular. 

When the manholes of small sewers are of necessity at long 
intervals, small shafts, large enough to admit of the lowering 
of a light of some description, are sometimes placed between 
manholes. These are called lamp-holes. 

Catch-basin.—As heretofore described, a catch-basin is a 
shallow well constructed near the curb of a street to catch the 
water which flows in the gutters, and re¬ 
move matters in suspension before discharg¬ 
ing the water into the sewers. A vertical 
section of such a basin through the inlet 
and outlet is shown in Fig. 180. The basin 
is constructed under the sidewalk adjacent 
to the curb. The water enters at C and 
flows into the sewer from A. The silt which 
collects at the bottom of the basin is removed 
through the manhole B. The partition between A and B holds 
the floating material and also forms a trap. 

Flow in Sewers.—The velocity of flow in a sewer should not 
be so small that deposits will be formed in the sewer, nor should 
the velocity be so great that stones will be swept along swiftly 
and chip the bottom of the sewer. The velocities recommended 
in Moore are: 

Minimum. Maximum. 

Sewers 6 to 9 inches. ... 3 feet per second 4 feet per second 

“ 12 to 24 “. 2* “ “ “ 4 “ “ 

“ large. 2 “ “ “ 4 “ “ 

These velocities are recommended for sewers either full or 
half-full. 

The corresponding slopes can be determined by solving the 
formula v=c\RS for S, substituting for v, c, and R their proper 
values. 

Variation of Velocity with Depth.—For any given circular 
sewer the values of c and S in the equation v=c v 7 RS are con¬ 
stant, since they depend only on the material of the walls and 
the slope of the sewer. In that case v will vary directly with 
■%/ R, the square root of the hydraulic mean depth, or the square 





















552 


CIVIL ENGINEERING. 


root of the ratio of the area of cross-section to its wetted pe¬ 
rimeter. 

Let A= area of cross-section of stream in a circular sewer 
in square feet; 

P = wetted perimeter or wetted arc of cross-section in feet; 
D = diameter of sewer in feet; 

cj) = angle subtended at the center of the circle by P in 
radius. 

Then we shall have 




D 

4 


D 2 sin </> 


D D 2 sin (j> 

4 4? 

sin <j>\ 

T)■ 


If we substitute for <j> different values and deduce the corre¬ 
sponding values of R, we may construct the following table: 


Values of R . 



R 

4 > 

R 


R 


D 


D 

260° 

D 

oo° 

.00— 

A 

140° 

,74 7 

I .21 — 

A 

20 

.02“ 

160 

t-: 

00 

00 

280 

1.20“ 

40 

.08“ 

180 

1.00“ 

3 °° 

1.16“ 

6o 

. 18“ 

200 

1.09' ‘ 

320 

1.11“ 

So 

• 30 “ 

220 

1.16“ 

340 

1.06“ 

IOO 

120 

• 44 “ 

• 59 “ 

240 

1.20“ 

360 

1.00“ 


The maximum value of R will be where the sin <f> is nega* 

sin <f> 


tive and the ratio 
Differentiating 


<t> 

sin <f> 

</> 

'sin <)) 


d 


is a numerical maximum, 
with respect to <f> we have 
CQS ^ -sin $ 


$ r ^ ^2 

Placing first differential coefficient equal to zero, 

(f> cos 4 >— sin <j> sin 6 

= o, or (/> = - - • ; L = tan <ft, 




COS (j) 





























SEWERAGE. 


553 


will give a maximum. R will therefore be a maximum when <j> 
equals about 257 degrees. 

From the table it is seen that the velocity, which varies directly 

with \ R, increases as the depth increases until the wetted pe¬ 
rimeter is 257 degrees or the depth is about 0.8D; the velocity 
will then decrease as the depth is still further increased. Ti e 
variation in the velocity as the wetted perimeter increases from 
zero to 180 degrees is very great, but the variation in the velocity 
as the wetted perimeter increases from 180 to 360 degrees is 
very slight. The velocity in a sewer one-fourth full is nearly 
0.8 of the velocity in a half-full sewer. 

Since the discharge is the product of the area of cross-section 
of a stream and its velocity, it follows from the table that the 
maximum discharge will take place when the sewer is almost 
but not quite full. 

Tables and curves of velocity and discharge for circular sewers 
running full, corresponding to different values of S and to different 
diameters, are given in works on sewerage. 

The variation in the value of R for egg-shaped sewers differs 
somewhat from that of the circular sewer; R and v are both 
greater for slight depths than in the circular sewer. Tables and 
curves of velocity for standard shapes of egg-shaped sewers are 
also given in works on sewerage. 

Area of Cross-section.—The required area of cross-section 
of a sewer is determined by the volume of sewage which it must 
carry, and by the maximum slope that may be given it. The 
volume of house-sewage is approximately equal to and never 
exceeds the amount of water introduced into the houses which 
are tributary to it. This volume will depend on the per-capita 
consumption and the density of the population. In all sewer 
systems allowance must be made for an increase in this density 
due to the growth of the community. 

The amount of factory sewage may be approximately deter¬ 
mined in the same manner. » 

The volume of storm-water which must be provided for in 
the construction of a sewer depends on the extent, slope, and 
character of the area tributary to the sewer, and on the maxi¬ 
mum rainfall. In a separate system the area tributary to the 
house-sewer may be nothing, it may be only the roof area, or it 


554 


CIVIL ENGINEERING. 


may be the roof and yard area. In a combined system it is the 
entire area whose surface-waters flow into the sewer, and comprises 
not only the roof and yard areas, but also the streets, parks, etc. 

Numerous attempts have been made to determine an empirical 
formula that will give the volume of water reaching the sewer 
in a unit of time. 

Among those most commonly employed is the Burkli-Ziegler 
formula given on page 498. In this country the maximum 
rainfall employed in that formula is often assumed as great as 
2.75 inches per hour. 

Flushing.—The velocity of flow in a sewer may not be suffi¬ 
cient to prevent the formation of deposits. Under such circum¬ 
stances it is necessary to flush the sewer at regular intervals by 
introducing into it a volume of water; if the slope is slight, a 
sufficient head must be given. Automatic flushing-tanks are 
usually basins which are gradually filled by a small pipe attached 
to a water-main, and discharged quickly by means of a large 
siphon-pipe. They are placed at the dead-ends of main sewers 
and branches. 

Ventilation.—Sewers are ventilated as much as possible to 
dilute the sewer-air and thus render it less offensive and nox¬ 
ious. The ordinary method is to provide perforated covers for 
the manholes and thus give entrance to the outer air. They 
may be ventilated through the soil-, waste-, and conductor-pipes 
of the houses by omitting the trap in the house-drain. They may 
also be ventilated by the hollow posts of street lights, either gas 
or electric; if a gas-jet burns at the top of these poles, it will pro¬ 
duce an artificial draft and aid in the ventilation. The sewer-gas 
may also be purified by burning disinfectants in the sewer itself. 

Sewage Disposal. 

Raw sewage, or the sewage as found in sewers, is a turbid 
liquid containing animal and vegetable organic matter, as well 
as inorganic matter, in suspension and solution. These matters 
are in varying proportions; their exact quantity and character 
can be determined only by chemical analysis. 

The organic matter in the presence of air and moisture is 
subject to putrefaction and reduction to simpler inorganic com¬ 
pounds and elements. During the process of reduction foul 


SEWERAGE. 


555 


odors are given off which, if not positively dangerous to human 
life, are at least sufficiently objectionable to constitute a nui¬ 
sance unless the process is properly regulated. 

The primary aim of every disposal scheme is to avoid the 
creation of such a nuisance. 

Raw sewage is distinguished from purified sewage by the 
absence of nitrates, by its large amount of ammonia, by the large 
amount of oxygen it will absorb, and by the large amount of 
matter in suspension. Purified sewage is distinguished from 
pure water by its large amount of chlorine; the chlorine in raw 
sewage is not altered in amount by any reduction process. 

Bacteria.—Raw sewage also contains myriads of microscopi: 
organisms called bacteria; several millions are often found in a 
cubic centimeter. Their function seems to be to break down the 
complex organic matters in the sewage and with the aid of oxy¬ 
gen to form simple stable compounds and elements. Pasteur, 
who established the fact that fermentation and putrefaction 
took place only in the presence of living organisms, divided these 
organisms into two classes, aerobes and anaerobes. The former 
live and work only in a medium well supplied with oxygen, while 
the latter live and work in a medium in which there is no oxy- 
gen. To these has since been added a third class, jacultative 
bacteria, which live and work in a medium in which oxygen is 
present, but in a small ratio to the other elements. As there 
are many varieties of each class and as each variety has its own 
mode of action, the successive steps in the reduction of the 
organic matter are extremely complicated. 

In general it may be said if sewage is spread intermittently 
in a thin sheet over a porous soil, such as coarse sand, whose 
pores are filled with air, the conditions are favorable for aerobic 
bacteria, -and the work will be done principally by them. The 
same is true if the sewage is discharged into highly aerated water. 
The effect of these bacteria being to oxidize the nitrogen of the 
organic matter and produce nitrates, the process of reduction 
by aerobic bacteria is called nitrification. It is a process of re¬ 
duction unaccompanied by foul odors. To prepare sewage for 
aerobic action it is usually screened to remove sticks, paper, rags, 
etc., not easily reduced, and is also allowed to stand for a few 
hours in settling-tanks to remove the coarser materials in suspen- 


55$ 


CIVIL ENGINEERING. 


sion not removed by the screen, which would be liable to close 
the pores of the filtering material. 

In general it may be said that if sewage stands in tanks or 
pools for any length of time, its oxygen is soon exhausted by the 
aerobes, and the conditions are then favorable for anaerobic action. 
The anaerobes break down the organic matter and form am¬ 
monia, nitrites, and release such gases as nitrogen, hydrogen, 
carbon dioxide, marsh-gas, and sulphuretted hydrogen. As 
nearly all the vegetable and animal organic matter is liquefied 
by the anaerobes, the process is called liquefaction or hydrolysis. 
During the process of liquefaction foul odors are given off, and 
the sewage itself is usually in a foul condition and requires the 
action of the aerobic bacteria to purify it. 

The process of reduction or purification to prevent a nui¬ 
sance is therefore usually effected by bacteria of all three varieties. 
If the sewage is screened and settled, the reduction, as explained 
hereafter, may be effected almost wholly by aerobic bacteria, but 
the other varieties will be present in some stage of the process 
and will assist. If the sewage is kept in an open or closed tank 
for twenty-four hours, the aerobic bacteria will almost disappear, 
and the process in the tank will be almost wholly anaerobic. In 
preparing the sewage for anaerobic action it is not so necessary 
to strain it, nor is it necessary to first pass it through a settling- 
tank. The anaerobic action must, however, always be followed 
by aerobic action. 

Bacteria may also be divided into the disease-producing 
bacteria and those that are not disease-producing. The princi¬ 
pal varieties of the first class, or pathogenic bacteria, which may 
be found in sewage, are those of cholera, diarrhoea, dysentery, 
and typhoid jever. They are found in the discharges of patients 
who have these diseases. If sewage containing such bacteria 
is discharged into any stream used as a water- or ice-supply, they 
at once become a menace to public health. 

Disposal.—Sewage, either in a raw state or after more or 
less screening and purification, may be disposed of by discharg¬ 
ing it into a body of water, upon natural land, upon prepared 
land, or upon specially constructed filter-beds. If the discharge 
is into fresh water, as is usually the case in inland cities, care must 
be taken not only to avoid a nuisance caused by too great a con- 


SEWERAGE. 


557 


centration of sewage or its deposition on the shores, but also to 
reduce the number of bacteria in it as much as possible, so that 
the pathogenic bacteria which may find access to it are destroyed. 

Dilution.—If raw sewage is discharged into a body of water 
simply to weaken the sewage and thus render it inoffensive, it is 
said to be disposed of by dilution. 

No nuisance will be created if raw sewage is discharged into 
a stream of moderate velocity whose dilution ratio, or the ratio 
of the volume of discharge of the stream to the sewage, is about 
forty. If the stream is a rapid-flowing mountain torrent, the factor 
may be reduced to twenty. These dilution ratios will vary some¬ 
what with the character of the sewage. When sewage is discharged 
into still water, such as a lake, the discharge should not be concen¬ 
trated at a single point, as it will then putrefy. Its distribution 
may be effected by discharging the sewage through a long con¬ 
duit, laid in the bed of the lake, with outlets at intervals along it. 

In still water the mixed water and sewage is so rapidly puri¬ 
fied by the process of sedimentation that the pollution of the 
water does not extend to a great distance from the sewer-outlet. 
At Burlington, Vt., a city of 15,000 inhabitants, which dis¬ 
charges its sewage at a single point into Lake Champlain, the 
presence of sewage cannot be detected by chemical processes at 
a distance of half a mile from the sewer-outlet. In running 
water the sedimentation is less rapid, and the pollution of the 
water is observed at a greater distance from the sewer-outlet. 
The reduction of the organic matter is effected by bacteria and 
other organisms which are found in the sewage, in the water, 
and in the mud of the bed. In the process of reduction oxygen 
is extracted from the water to form the new compounds. As the 
organisms which effect the final change are aerobic, this process 
is stopped, and offensive decomposition begins whenever the oxy¬ 
gen in the water is too greatly reduced. It is for this reason that 
the discharge into still water should be distributed over a wide 
area, and that a stream which can purify a given volume of sew¬ 
age becomes foul when charged with a greater one. It ex¬ 
plains why a torrent which is constantly aerated can have a smaller 
dilution ratio than a slow stream of equal volume. 

The bacteria of the sewage are reduced largely by the pro¬ 
cess of sedimentation which carries them to the bottom with the 


558 


CIVIL ENGINEERING. 


suspended material. Other causes of purification are the lack 
of food and the unfavorable surroundings in the clarified water. 
The reduction of the number of these organisms in still water 
is very great. At Burlington, Vt., they are said to be reduced 
from one million to one thousand per cubic centimeter in one 
hundred feet. Judging from the water-borne diseases among 
the consumers of the water-supply, it was assumed, however, 
that the bacteria were not wholly removed at a distance of a 
half-mile, the original distance between the outlet-sewer and the 
inlet of the water-supply in Lake Champlain. 

In running streams bacteria must be carried to considerable 
distances, since it has been shown by experiments that the bac¬ 
teria of typhoid fever may live twenty-four days even in ice-cold 
water. 

From the above it would appear that raw sewage may be 
disposed of by dilution without creating a nuisance if the volume 
of running water or the area of distribution in still water is suffi¬ 
ciently great. It will, however, always be more or less danger¬ 
ous to discharge it into waters which may be used for water- and 
ice-supplies on account of the pathogenic bacteria it may con¬ 
tain. As the volume of discharge of most inland streams is 
small, and as they are in addition liable to be used as water- and 
ice-supplies, efforts are being made everywhere to prevent the 
discharge of raw sewage into small fresh-water streams or ponds 
without previous treatment to reduce the organic matter and 
the number of bacteria. 

If the discharge is into salt water, which cannot be em¬ 
ployed for domestic use unless it is distilled, nc possible harm 
can be done except to marine life. The sewage, however, should 
be screened of all floating materials, and then be discharged, 
below the surface into a current which will carry it away from 
the shores and prevent deposits which will be exposed at low 
tides. To prevent the latter it is often necessary to construct 
reservoirs to hold the sewage which reaches the outlet during 
flood-tide, and discharge it only when the tide is at ebb. Such 
a reservoir is an element of the Boston sewerage system. The 
principal danger resulting from the discharge of raw sewage into 
salt water is the infection of oysters by pathogenic bacteria. 
Cases of infection thus propagated are of record. 


SEWERAGE. 


559 


Broad Irrigation.—Disposal by irrigation consists in apply¬ 
ing screened sewage to the growing vegetation of a sewage farm. 
The greater part of the liquid is absorbed by the vegetation, and 
the remainder after filtration through the soil may be caught in 
subsoil drains and conveyed to the natural drainage streams of 
the country. The solids not removed by previous screening are 
absorbed by the soil in the same manner as manures and fer¬ 
tilizers applied to land. The main objection to irrigation is the 
great extent of land required; about an acre is required for every 
25 to 100 people who contribute to the sewage, depending on 
the character of the soil, and the irregularity of the sewage-supply 
due to rainfall. This method of disposal is, however, extensively 
employed, especially in Europe. 

In level country the sewage may be conveyed in parallel 
troughs about 40 feet apart, raised slightly above the surface of 
the soil. From these it overflows the ground, properly sloped, on 
either side, and the surplus is carried off in surface or subsoil 
drains midway between the troughs. Another method is to shape 
the ground in alternate ridges and furrows and let the sewage flow 
in the furrows. It thus reaches only the roots of the plants which 
grow on the ridges; subsoil drains may be placed beneath the 
ridges. As the nitrogen of the organic material must be reduced 
to nitrates before it can be absorbed by plants, the bacterial action 
in irrigation must be wholly aerobic if it is desired to avoid foul 
odors. This necessitates a porous soil and the even distribution 
of sewage; the formation of pools must be avoided. The sewage 
must be applied intermittently if the pores of the soil show any 
signs of becoming clogged. 

Intermittent Filtration.—This is a modified form of irriga¬ 
tion in which the sewage is applied to the land, not for the pur¬ 
pose of utilizing the sewage for plants, but rather to purify as 
much sewage as possible per acre of land. The sewage is ap¬ 
plied intermittently with the view of supporting as large a num¬ 
ber of aerobic bacteria as possible and thus avoid the expense of 
large irrigating farms. The land used for intermittent filtra¬ 
tion requires more thorough underdrainage than that used for 
irrigation. The underdrains assist in ventilating the soil. 

The sewage is usually prepared by screening and then by 
standing for a few hours in settling-basins. The land is divided 


CIVIL ENGINEERING. 


c6o 

into a number of beds, some of which are settling-basins and the 
others are purifying-beds. The sewage is first conveyed to the 
settling-basins, where the sludge is precipitated by gravity, and 
from them to the purifiers. From time to time the settling- 
basins are emptied and the sludge dried by evaporation. When 
the sludge dries it is raked up and carted away to be burned, 
buried, or otherwise disposed of. There are always sufficient 
beds to allow each purifying-bed to rest dry some time after it is 
emptied. This process is extensively employed in this country in 
• regions having a sandy or gravelly soil. 

The natural soil may be replaced by specially constructed 
filter-beds of sand and gravel or other material thoroughly under¬ 
drained. The area of such beds will naturally be less than the 
area of natural soil required for purification. 

Chemical Precipitation.—The suspended solids which remain 
after screening are often removed by precipitating them by some 
chemical, as lime, copperas, or alum. The substances may be 
employed alone or with each other. The kind and the amount 
of the precipitate best suited to the sewage must be determined 
by analysis of the effluent. 

The chemicals are usually dissolved in water in a separate 
tank and then allowed to mix well with the sewage before the 
latter is admitted into the precipitation-tank. The process 
of precipitation may be intermittent or continuous. In the 
intermittent process there are three tanks, one being filled, one 
standing full, and the third being emptied. In the continuous 
tank the sewage charged with the precipitate flows slowly through 
the tank either in a horizontal or a vertical direction, depositing 
its sludge on the bottom. One of the favorite forms is the verti¬ 
cal, or Dortmund, tank. This is composed of two cylindrical 
concentric tubes of equal length, the diameter of the outer being 
about five times that of the inner. To the bottom of the outer 
tube is riveted an extension in the form of an inverted cone whose 
smaller end is equal to the diameter of the inner tube. The 
sewage charged with precipitate is admitted at the top of the 
inner tube, flows down to its bottom, where it is distributed by 
radial arms through the cross-section of the larger one. It then 
flows up through the outer tube and is discharged at its top. 
The sludge sinks to the bottom and is collected at the bottom 


SEWERAGE. 561 

of the conical extension of the outer tube; from there it is dis¬ 
charged through a discharge-pipe by gravity or by pumping. 

If water transportation is available, the sludge is usually 
received in closed boats and carried out to sea. Otherwise it 
must be discharged into settling-basins, where it is allowed to 
evaporate. When nearly dry it is molded into cakes and burned. 

The effluent of precipitation-tanks is clear liquid, but unless 
the process also removes the dissolved organic substances it 
is subject to further decomposition and cannot be discharged 
into streams without creating a nuisance. The effluent also 
usually contains a larger number of bacteria than is considered 
safe for its discharge into streams which are employed as water- 
or ice-supplies. 

Land Required.—The amount of land required for the above 
processes has been determined in England to be as follows: 



Soil. 

Number 
of Acres. 

Number of 
Persons. 

Simple irrigation. j 

Simole intermittent filtration. 

Stiff clay 
Loamy gravel 
Sandy gravel 
Clay 

Loamy gravel 

Sandy gravel 

I 

I 

I 

I 

I 

I 

I 

25 

IOO 

100 to 300 
200 

400 

500 to 600 

2000 

simple mtermitte . . . f 

Irrigation preceded by precipitation. < 

Intermittent filtration preceded by 
r>rprini fa f ion . 

Intermittent filtration in specially 
prepared filters, preceded by pre- 
+ o fiir^n and fnllnwpH lw irH era f iotl 




Contact-beds.— Contact-beds are specially prepared filter-beds 
arranged in series. They were first installed in Sutton, England, 
to replace a system of disposal by precipitation and irrigation 
which had proved unsatisfactory. 

The settling-tanks, which were about 30 by 50 feet in area, 
were converted into coarse-contact beds by filling them to a depth 
of 3 J feet with burned-clay ballast, or clay burned and reduced to 
1-inch fragments. It was drained by a 6-inch underdrain to 
which were attached, at intervals of 3 feet, parallel branches 3 
inches in diameter. The sewage was fed to the bed through 
a trough supported above it. 

The -fine-contact beds were each about 20 by 40 feet, and were 
filled to a depth of 3 feet by various combinations of material 


















562 


CIVIL ENGINEERING. 


screened to pass through a f-inch screen. These materials were 
burned clay, coke breeze, gravel, and different varieties of sand. 
The sewage was fed to these beds in the same manner as to the 
coarse ones. 

In operating the beds, screened sewage was first allowed to flow 
on a coarse bed, whose outlet was closed, until the bed was filled 
to within 6 inches of the surface. The bed was then allowed to 
remain full for two hours. At the end of that time it was emptied, 
and the effluent was distributed slowly over one or more of the 
fine beds; this operation took about an hour. The effluent of 
the fine beds was sufficiently purified to admit its discharge into 
a stream which flowed to the sewage farm. The beds were 
allowed to stand empty for at least two hours before being again 
used. As a rule the intervals between the filling of each coarse 
bed was eight hours. 

While porous material like coke is the best for contact-beds, 
especially for the fine ones, beds have been satisfactorily worked 
with broken stone, slag, coal, etc. 

In some of the later beds a series of three beds instead of 
two have been employed, coarse, medium, and fine; the effluent 
of such a series is purer than from a series of two. 

To more thoroughly aerate the sewage and the filter-beds, 
a revolving sprinkler has been devised like that employed in 
sprinkling lawns. Where this sprinkler is employed the beds 
are circular. 

In the contact-beds as above described the action is princi¬ 
pally aerobic, although there must be some anaerobic action in 
the coarse bed; it is, however, not sufficient to produce disagree¬ 
able odors. The aerobic action in the coarse bed may be in¬ 
creased by employing a revolving sprinkler to distribute the 
sewage over it. 

Most of the sludge is destroyed by bacterial action, and this 
is one of the advantages of the system. This requires the cleans¬ 
ing of the beds only at long intervals. 

Septic Tanks.—The septic tank is an open or closed tank 
in which raw sewage is allowed to stand for a period of about 
twenty-four hours for the purpose of liquefying its organic matter 
by anaerobic action. It differs from a settling-tank only in the 
length of time the sewage remains in it. In this tank the sewage 


SEWERAGE. 


5 6 S 


putrefies, such gases as hydrogen, nitrogen, marsh-gas, carbon 
dioxide, and hydrogen sulphide are given off, and ammonia is 
formed. A thick scum forms over the top of the tank, and the 

sludge is deposited in a fine powder. On account of its odors 

$ 

a septic tank usually creates a nuisance if placed near dwellings. 

Although the amount of sludge is much less than in the simple 
settling-tank and the tank therefore requires less frequent clean¬ 
ing, the sludge produced is of a more offensive character. 

The effluent of a septic tank usually requires further puri¬ 
fication by passing through fine filter-beds like those of the con¬ 
tact system. Before it reaches the fine filters it is aerated by 
passing in a thin sheet over an aerating-weir. The weir and 
beds produce nitrification, which is the last stage of every system 
of sewage treatment. 

Septic tanks have been introduced in this country where 
the factory wastes can be reduced only by anaerobic action, and 
where the soil is clayey and not suited to intermittent filtration. 
The tanks are either open or closed; the latter are best in very 
cold climates. The sewage is admitted some distance below 
the level of the liquid in the tank, and usually into a grit-cham¬ 
ber separated from the main tank by a low wall over which the 
sewage flows. It is also withdrawn from the tank by a pipe 
tapping the tank some distance below the surface of the liquid. 
The most suitable rate of flow through the tank is determined 
by experiment. 

A modified form of septic tank is a coarse-contact bed in which 
the sewage is admitted at the bottom and escapes to the fine bed 
from the top. This is called upward filtration. 

Destruction of Bacteria.—Bacteria are removed from sew¬ 
age either by filtering through the soil, as in the methods of irriga¬ 
tion and intermittent filtration, or by passing through the fine 
beds of contact-filters and septic tanks. 

For further information see Gerhard’s “Sanitary Engineering of Build¬ 
ings,” Folwell’s “Sewerage,” Moore’s “Sanitary Engineering,” Sedgwick’s 
“Principles of Sanitary Science,” Rideal’s “Sewage and the Bacterial Puri¬ 
fication of Sewage,” Baumeister’s “Cleaning and Sewerage of Cities,” and 
Rafter and Baker’s “ Sewage Disposal in the United States.” 



























' 

• , 


































' 
























I 





















































INDEX. 


PAGE 

Arch, abutments and unsymmetrical loading. 291 

allowable masonry and pressures. 289 

bridges. 452 

conditions of stability. 284 

crown-thrust. 281 

definition, parts and kinds. 277 

depth at keystone and springing-line. 288 

elastic. 291 

load on an. 287 

Mery’s curve of pressure. 285 

modes of rupture. 279 

testing the design of. 289 

Asphalt. 386 

Beam, fixed at end, definition. 69 

strongest from circular log. 65 

Beams, continuous. 90 

of uniform cross-section. 54, 63 

rolled 1 . 175 

solid built. 174 

stiffness of. 86 

strength of. 86 

of uniform strength. 55, 56 

Brick, classification and tests. 369 

glazed, sewer-pipe. 37r 

manufacture. 368 

paving-, fire-, terra-cotta, tiles. 370, 371 

Bridge, approaches and abutments. 445 

cantilever . 45 ° 

camber. 44 ^ 

definition of different parts. 442 

draw-. 469 

erection. 46S 

floors. 44 b 

5 6 5 




































566 


INDEX. 


PAGE 

Bridge, Howe truss. 451 

loads on a highway. 462 

“ “ “ railway. 464 

masonry arch. 452 

piers. 444 

plate girder. 448 

reinforced concrete. 455 

steel arch. 455 

suspension. 457 

temperature stresses in a. 468 

truss. 448 

wind-bracing of a. 465 

wind pressure on a. 465 

Brooklyn bridge. 439, 461, 471 

Caisson, common. 428 

pneumatic . 435 

“ of New York and Forth bridges. 439 

Cement, classification. 374 

manufacture. 375 

preservation. 376 

tests. 377 

Coffer-dams. 433 

Columns, built-up. 150 

definitions. 141 

design. 152 

Euler’s formulas. 141 

Gordon’s formulas. 144, 147 

Rankine’s formulas. 144, 148 

right line formulas. 149 

Concrete, asphalt and coal-tar. 386 

cement manufacture. 384 

“ use and strength. 385 

masonry. 399 

reinforced. 401 

“ beams. 405 

“ columns. 403 

“ systems. 412 

Counterbraces in a truss. 221 

Dam, curve of pressure. 266 

definitions and parts. 260 

profile of a, conditions governing... 262 

“ “ , economical. ..... 262 

“ “ “ , high. 266 

water pressure on a. 260 















































INDEX. 


5 6 7 


PAGE 

Deflections in beams, table of values. 86 

“ , maximum allowable. 87 

Derrick, boom. 256 

Elasticity, general laws of. 8 

in compression. 12 

“ flexure. 70 

“ shear. 32 

“ tension. 9 

“ torsion. 39 

coefficient of lateral. 40 

“ “ longitudinal. 14 

limit of longitudinal. 14 

Elongation of a rod. 23 

Equilibrium, conditions and equations of. 1 

Eye-bars. 167 

Factors of safety, columns. 148 

general values. 52 

Fiber, mean, of beam fixed at both ends. 76, 78 

“ “ “ “one end. 80,84 

“ “ resting on end supports. 74, 75 

“ cantilever beam. 71, 73 

Flexure, theory of. 47 

Floors, I-beam. 477 

loads on. 479 

reinforced concrete. 478 

wooden. 476 

Forth bridge. 44 °> 45 °» 467* 47 1 

Foundations, bearing power of soils. 416 

general principles governing. 414 

in firm soils. 4 X 7 

“ soft “ . 418 

pile. 4 2 5 

random rock. 426 

under water. 4 2 5 

Graphic method, bending moment. 243 

bending moment on a pin. 246 

center of gravity. 242 

equilibrium polygon. 230 

l ' “ through two and three points. 249 

force polygon. 226 

moment of inertia. 248 

nomenclature, -. 226 











































5 68 


INDEX. 


PAGE 


Graphic method, position diagram. 226 

representation of forces. 225 

solution of concurrent forces. 225 

“ “ non-current forces. 228 

“ “ a simple roof-truss. 235 

Gravel. 3& 1 


Highways, construction of. 

cross-section. 

definitions. 

drainage. 

earth roads. 

French. 

grades. 

gravel roads. 

locating and surveying. . 

macadam roads. 

resistance of vehicles. . . . 

tel ford roads. 

tractive power of a horse 
Hawksbury bridge. 


••• 495 
... 487 

... 481 
... 497 
... 502 
... 488 
... 486 

••• 5°3 
... 490 

5 ° 4 , 5 oS 
... 483 
... 507 
..i 482 
... 432 


Impact force, in flexure. 88 

longitudinal. 21 

Iron, cast, defects, durability, and preservation. 350 

malleable. . .. 352 

manufacture and properties. 349 

specifications and tests. 358 

Ironwork, joints in. 359 

Iron, wrought, defects and preservation. 358 

manufacture and properties. 357 


Lacing and latticing of a column. 

Lime, common, characteristics and manufacture. , 

hydraulic, characteristics. 

preservation and tests. 

Live load, effect of. 

“ on a truss with parallel chords 
Luxemburg arch bridge. 


••• 153 
... 372 

••• 373 

••• 373 
... 119 

213-220 
••• 454 


Masonry, brick, allowable pressure on, retaining-walls, and arches 

bond and walls. 

concrete, allowable pressures. 

construction of walls. 

retaining-walls and arches. 

stone, allowable compression on. 


398 

397 

401 

399 
401 

392 












































INDEX. 


569 


PAGE 

Masonry, stone, arches and ovals. 302 

“ bond. 388 

“ pointing, settling. 395 

“ Rankine’s rules for. 387 

“ rubble and ashlar. 390 

“ settling and temperature effects. 396 

Modulus of elasticity. See Elasticity, crushing, definition and values.... 14 

flexure, definition and values. 51, 52 

shearing, definition and values. 33 

tenacity, definition. 12 

“ table of values. 14 

torsion, definition and values. 40 

longitudinal resilience, definition. 20 

section, definition and values. 52, 53 

Moment, bending. 45 

curve of, beam fixed at ends. 77, 78 

“ “ “ “ at one end . ... 83, 85 

“ “ “ resting on end supports. 59, 6° 

“ “ cantilever. 57, 58 

“ “ definition. 57 

due to live load on a beam. 119 

maximum, table of values. 86 

torsional. 36 

Moments, theorem of three. 90 

Mortar, cement. 3 8 3 

lime. 3 82 

uses of. 3 8 3 

Niagara River steel-arch bridges. 456, 47 1 

Omaha swing draw-bridge. 47 1 

Pavements, asphalt. 5 J 6 

brick. 5 i 4 

cobblestone. 5 X 3 

foundations. 5°9 

stone block. 5 12 

wearing surfaces. 5 ° 8 

wood. 5 10 

Piers, crib. 444 

masonry. 444 

Piles, allowable loads on. 422 

definitions. 4 l8 

Pile-driving machinery .. 420 

Plate girder, design of a. I7<5 ’ 190 

TV ... 166 

Pin-joint. 












































57° 


INDEX. 


PAGE 

Pins, description. 168 

design. 168 

Pipes, determination of thickness. 28 

flow of water through. 303 

sewer. 548 

water-mains. 539 

Resilience, in flexure. 87 

longitudinal. 19 

torsional. 42 

Retaining-walls, conditions governing profile of. 273 

Coulomb’s formula for thickness. 271 

counterforts and buttresses. 275 

empirical formulas for thickness. 274 

foundations and drainage. 275 

Poncelet’s formula for thickness. 272 

Rankine’s formula for thickness. 273 

Riveted joints, description of. 158 

design of. 160 

modes of yielding. 160 

Rivets, description of. 158 

Roofs, description of parts and varieties. 472 

loads. 474 

methods of construction. 473 

wind pressure on. 474 

Sand. 380 

Sewage, definition. 542 

Sewage disposal, bacteria. 555 

contact-beds. 561 

destruction of bacteria. 563 

filtration. 559 

irrigation. 559 

methods. 556 

precipitation. 360 

septic tanks. 362 

Sewerage, combined system. 342 

definitions. 342 

house drainage. 343 

household wastes. 343 

separate system. 342 

tra P s . 545 

Sewers, area of cross-section. 333 

brick and concrete. 349 

catch-basins. 33! 













































INDEX. 


5 7i 


PAGE 

Sewers, flow in. ^51 

flushing. 554 

lamp-holes. 551 

manholes. 550 

Pipe. 548 

principles governing construction. 547 

ventilation. 554 

Shafts, power transmitted by. 43 

Shear, curve of, in beam fixed at both ends. 78 

“ “ “ “ one end. 83, 85 

“ “ resting on end supports. 60, 61 

“ cantilever beam. 58 

horizontal, in a beam. 98 

in a beam due to a live load. 129 

“ “ rod or column. 34 

section of zero shear in a beam. 137 

simple. 32 

table of safe and ultimate unit stresses in. 33 

unit stresses in planes perpendicular to each other. 35 

vertical shear in a horizontal beam. 46 

Spheres, resistance of, to internal pressure. 30 

Steel, durability, specifications, and tests. 355 

manufacture and classification.-. 352 

properties. 353 

strength and defects. 354 

Stone, artificial cement. 385 

broken. 381 

natural, durability and preservation. 364 

“ mechanical properties. 363 

“ physical properties. 362 

“ quarrying and dressing. 365 

“ specifications and tests. 367 

“ structure and classification. 361 

Strain, definition of. 7 

Stress, definition of. 7 

safe unit, definition of. 14 

“ “ values in compression and tension. 14 

“ “ “ “ flexure. 52 

“ “ “ “ shear. 33 

“ “ “ “ torsion. 41 

under eccentric loading. no 

Stresses, classification of. 7 

combined. 103 

due to changes of temperature. 30 

table of maximum stresses in beams. 86 

Structure, engineering, definition of. 1 
















































572 


INDEX. 


PAGE 

Structural design, operations in. 2 

Suspension bridge. 457 

Timber construction, joints in. 341 

defects in. 334 

durability and preservation. 335 

market forms and designations. 339 

mechanical properties. 332 

physical properties. 330 

trees, classification and structure. 326 

Topeka bridge. 455 

Torsion, safe and ultimate stresses in. 40 

theory of. 36 

Towers, steel bridge. 444 

water. 539 

Trestles, steel. 444 

wooden. 442 

Truss, compound. 254 

definition of a. 191 

determination of stresses in, by analysis of systems of concurrent 

forces. 193 

by analysis of moments. 203 

by method of sections. 210 

loading. 224 

queen-post, stresses in a. 222 

three-hinged. 255 

Trusses, Baltimore, cantilever, and Petit. 251 

bowstring and continuous. 252 

classification of. 191 

Howe, Pratt, and Warren. 192 

Howe. 451 


Uniform strength, beam of. 55 

shaft of. 42 

rod of. 25 

Viaur bridge or viaduct. 456, 467, 471 

Water, discharge through an orifice. 

short tubes. . . . 

“ “ weirs. 

flow in open channels.. 

“ “ pipes of varying diameters 

“ through pipes. 

“ “ “ formulas for. . . 


297 

299 

3° 1 
3i8 
316 
303 
309 










































INDEX. 57 3 

PAGE 

Water, hydraulic grade line. 312 

loss of head at entrance. 303 

“ “ “ due to contraction. 316 

“ “ “ “ “ enlargement. 315 

“ “ “ “ “ friction. 305 

“ “ “ “ “ valves. 3x6 

measurement of discharge of rivers. 324 

miner’s inch. 301 

physical properties. 292 

static principles. 292 

velocity-head. 308 

velocity in open channels. 323 

W ater-supply, amount required. 521 

blow-offs and air-escape valves. 563 

character of. 520 

conduits and pumps. 531 

distributing-reservoirs. 538 

filters, rapid. 537 

“ slow.534 

measurement of. 522 

purifying system. 532 

receiving-reservoirs.. 5 2 5 

reservoir dams. 5 2 9 

run-off. 5 2 3 

settling-basins. 533 

spillways. 5 2 & 

springs and wells. 5 2 4 

water-mains. 539 

water-meters. 5^3 

Weights of engineering materials, table of. 5 

Well or open dredging process. 4 2 9 

Williamsburg bridge. 355 > 439 > 461, 47 1 

Work done by longitudinal forces. 














































. 
















. 







* 












































' 

































































short-title catalogue 

OF THE 

PUBLICATIONS 

OF 

JOHN WILEY & SONS, 

New York. 

London: CHAPMAN & HALL, Limited. 


ARRANGED UNDER SUBJECTS. 


Descriptive circulars sent on application. Books marked with an asterisk (*) are sold 
at net prices only, a double asterisk (**) books sold under the rules of the American 
Publishers’ Association at net prices subject to an extra charge for postage. All books 
are bound in cloth unless otherwise stated. 


AGRICULTURE. 

Armsby’s Manual of Cattle-feeding.nmo, $i 75 

Principles of Animal Nutrition.8vo, 4 00 

Budd and Hansen's American Horticultural Manual: 

Parti. Propagation, Culture, and Improvement.nmo, 1 50 

Part II. Systematic Pomology.nmo, 1 50 

Downing’s Fruits and Fruit-trees of America.8vo, 5 00 

Elliott’s Engineering for Land Drainage.nmo, 1 50 

Practical Farm Drainage.nmo, 1 00 

Green’s Principles of American Forestry.nmo, 1 50 

Grotenfelt’s Principles of Modern Dairy Practice. (Woll.).nmo, 2 00 

Kemp’s Landscape Gardening.nmo, 2 50 

Maynard’s Landscape Gardening as Applied to Home Decoration.nmo, 1 *0 

Sanderson’s Insects Injurious to Staple Crops.i2mo, 1 50 

Insects Injurious to Garden Crops. (In preparation.) 

Insects Injuring Fruits. (In preparation.) 

Stockbridge’s Rocks and Soils. 8vo, 2 50 

Woll’s Handbook for Farmers and Dairymen.i6mo, 1 50 

ARCHITECTURE. 

Baldwin’s Steam Heating for Buildings.i2mo, 2 50 

Berg’s Buildings and Structures of American Railroads.4to, 5 00 

Birkmire’s Planning and Construction of American Theatres.8vo, 3 00 

Architectural Iron and Steel.8vo, 3 50 

Compound Riveted Girders as Applied in Buildings.8vo, 2 00 

P lanni ng and Construction of High Office Buildings.8vo 3 50 

Skeleton Construction in Buildings.8vo, 3 00 

Brigg’s Modern American School Buildings. 8vo, 4 00 

Carpenter’s Heating and Ventilating of Buildings.8vo, 4 00 

Freitag's Architectural Engineering.8vo, 3 50 

Fireproofing of Steel Buildings.8vo, 2 50 

French and Ives’s Stereotomy.8vo, 2 50 

Gerhard’s Guide to Sanitary House-inspection.i6mo, 1 00 

Theatre Fires and Panics..i2mo, 1 50 

Holly’s Carpenters’ and Joiners’ Handbook.i8mo, 75 

Johnson’S Statics by Algebraic and Graphic Methods.8vo, 2 00 


































Kidder’s Architects’ and Builders’ Pocket-book. Rewritten Edition. i6mo, mor., 


Merrill’s Stones for Building and Decoration.8vo, 

Non-metallic Minerals: Their Occurrence and Uses.8vo, 

Monckton’s Stair-building.4to, 

Patton’s Practical Treatise on Foundations.8vo, 

Peabody’s Naval Architecture. . . ..8vo, 

Richey’s Handbook for Superintendents of Construction.i6mo, mor , 

Sabin’s Industrial and Artistic Technology of Paints and Varnish.8vo, 

Siebert and Biggin’s Modern Stone-cutting and Masonry.8vo, 

Snow’s Principal Species of Wood.8vo, 

Sondericker’s Graphic Statics with Applications to Trusses, Beams, and Arches. 

8vo, 

Towne’s Locks and Builders’ Hardware.i8mo, morocco, 

Wait’s Engineering and Architectural Jurisprudence.8vo, 

Sheep, 

Law of Operations Preliminary to Construction in Engineering and Archi¬ 
tecture.8vo, 

Sheep, 

Law of Contracts.8vo, 

Wood’s Rustless Coatings: Corrosion and Electrolysis of Iron and Steel. .8vo, 

Woodbury’s F«re Protection of Mills. 8vo, 

Worcester and Atkinson’s Small Hospitals, Establishment and Maintenance, 
Suggestions for Hospital Architecture, with Plans for a Small Hospital. 

i2mo, 

The World’s Columbian Exposition of 1893.Large 4to, 


.5 

5 

4 

4 

5 
7 

4 
3 

1 

3 

2 

3 

6 
6 

5 
5 

3 

4 
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00 

00 

50 

00 

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50 

50 

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50 

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50 


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ARMY AND NAVY. 


Bernadou’s Smokeless Powder, Nitro-cellulose, and the Theory of the Cellulose 


Molecule..i2mo, 

* Bruff’s Text-book Ordnance and Gunnery.8vo, 

Chase’s Screw Propellers and Marine Propulsion.8vo, 

Cloke’s Gunner’s Examiner. (In press.) 

Craig’s Azimuth.4to, 

Crehore and Squier’s Polarizing Photo-chronograph.8vo, 

Cronkhite’s Gunnery for Non-commissioned Officers.24mo, morocco, 

* Davis’s Elements of Law.8vo, 

* Treatise on the Military Law of United States.8vo, 

Sheep, 

De Brack’s Cavalry Outposts Duties. (Carr.).241110, morocco, 

Dietz’s Soldier’s First Aid Handbook.i6mo, morocco, 

♦Dredge’s Modern French Artillery.4to, half morocco, 

Durand’s Resistance and Propulsion of Ships.8vo, 

* Dyer’s Handbook of Light Artillery.i2mo, 

Eissler’s Modern High Explosives.8vo, 

* Fiebeger’s Text-book on Field Fortification.Small 8vo, 

Hamilton’s The Gunner’s Catechism.i8mo, 

* Hoff’s Elementary Naval Tactics.8vo, 

Ingalls’s Handbook of Problems in Direct Fire.8vo, 

* Ballistic Tables.8vo, 

* Lyons’s Treatise on Electromagnetic Phenomena. Vols. I. and II. .8vo, each, 

* Mahan’s Permanent Fortifications. (Mercur.).8vo, half morocco, 

Manual for Courts-martial.i6mo, morocco, 

* Mercur’s Attack of Fortified Places.nmo, 

* Elements of the Art of War.8vo, 

Metcalf’s Cost of Manufactures—And the Administration of Workshops. .8vo, 

* Ordnance and Gunnery. 2 vols.i2mo, 

Murray’s Infantry Drill Regulations.i8mo, paper, 

Nixon’s Adjutants’ Manual.24mo, 

Peabody’s Naval Architecture.8vo, 


2 
6 

3 

3 

3 

2 

2 

7 

7 

2 
x 

15 

5 

3 

4 
2 
I 
X 

4 

1 

6 

7 

1 

2 

4 

5 
5 

1 

7 


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00 

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* Phelps's Practical Marine Surveying.g vo> 2 

Powell’s Army Officer’s Examiner. i2mo, 4 00 

Sharpe’s Art of Subsisting Armies in War.i8mo. morocco, 1 50 

* Walke’s Lectures on Explosives.8vo, 4 00 

* Wheeler’s Siege Operations and Military Mining.8vo, 2 00 

Winthrop’s Abridgment of Military Law.i2mo, 2 50 

Woodhull’s Notes on Military Hygiene. i6mo, 1 50 

Young’s Simple Elements of Navigation.i6mo, morocco, 1 00 

Second Edition, Enlarged and Revised.i6mo, morocco, 2 00 

ASSAYING. 


Fletcher’s Practical Instructions in Quantitative Assaying with the Blowpipe. 

i2mo, morocco, 1 50 


Furman’s Manual of Practical Assaying.8vo, 3 00 

Lodge’s Notes on Assaying and Metallurgical Laboratory Experiments. . . .8vo, 3 00 

Miller’s Manual of Assaying.i2mo, 1 00 

O’Driscoll’s Notes on the Treatment of Gold Ores.8vo, 2 00 

Ricketts and Miller’s Notes on Assaying.8vo, 3 00 

Ulke’s Modern Electrolytic Copper Refining.8vo, 3 00 

Wilson’s Cyanide Processes.i2mo, 1 50 

Chlorination Process.12mo, 1 50 

ASTRONOMY. 

Comstock’s Field Astronomy for Engineers.8vo, 2 50 

Craig’s Azimuth. 4 to, 350 

Doolittle’s Treatise on Practical Astronomy.8vo, 4 00 

Gore’s Elements of Geodesy.'..8vo, 2 50 

Hayford’s Text-book of Geodetic Astronomy.8vo, 3 00 

Merriman’s Elements of Precise Surveying and Geodesy.8vo, 2 50 

* Michie and Harlow’s Practical Astronomy.8vo, 3 00 

* White’s Elements of Theoretical and Descriptive Astronomy.i2mo, 2 00 

BOTANY. 


Davenport’s Statistical Methods, with Special Reference to Biological Variation. 

i6mo, morocco, 1 25 


Thome' and Bennett’s Structural and Physiological Botany.i6mo, 2 25 

Westermaier’s Compendium of General Botany. (Schneider.).8vo, 2 00 

CHEMISTRY. 

Adriance’s Laboratory Calculations and Specific Gravity Tables.nmo, 1 25 

Allen’s Tables for Iron Analysis.8vo, 3 00 

Arnold’s Compendium of Chemistry. (Mandel.).Small 8vo, 3 50 

Austen’s Notes for Chemical Students.nmo, 1 50 

Bernadou’s Smokeless Powder.—Nitro-cellulose, and Theory of the Cellulose 

Molecule.nmo, 2 50 

Bolton’s Quantitative Analysis.8vo, 1 50 

* Browning’s Introduction to the Rarer Elements.8vo, 1 50 

Brush and Penfield’s Manual of Determinative Mineralogy. 8vo, 4 00 

Classen’s Quantitative Chemical Analysis by Electrolysis. (Boltwood.). .8vo, 3 00 

Cohn’s Indicators and Test-papers.nmo, 2 00 

Tests and Reagents.8vo, 3 00 

Crafts’s Short Course in Qualitative Chemical Analysis. (Schaeffer.). .. nmo, 1 50 

Dolezalek’s Theory of the Lead Accumulator (Storage Battery). (Von 

Ende.).nmo, 2 50 

Drechsel’s Chemical Reactions. (Merrill.).nmo, 1 25 

Duhem’s Thermodynamics and Chemistry. (Eurgess.).8vo, 400 

Eissler’s Modern High Explosives.8vo, 4 00 

EfTront’s Enzymes and their Applications. (Prescott.).8vo, 300 

Erdmann’s Introduction to Chemical Preparations. (Dunlap.).nmo, 1 25 

3 













































Fletcher’s Practical Instructions in Quantitative Assaying with the Blowpipe. 

i2mo, morocco, 

Fowler’s Sewage Works Analyses..nmo, 

Fresenius’s Manual of Qualitative Chemical Analysis. (Wells.).8vo, 

Manual of Qualitative Chemical Analysis. Part I. Descriptive. (Wells.) 8vo, 

System of Instruction in Quantitative Chemical Analysis. (Cohn.) 
2 vols.8vo, 

Fuertes’s Water and Public Health.nmo, 

Furman’s Manual of Practical Assaying.8vo, 

* Getman’s Exercises in Physical Chemistry.X2mo, 

Gill’s Gas and Fuel Analysis for Engineers.nmo, 

Grotenfelt’s Principles of Modern Dairy Practice. (Woll.).i2mo, 

Hammarsten’s Text-book of Physiological Chemistry. (Mandel.).8vo, 

Helm’s Principles of Mathematical Chemistry. (Morgan.).nmo, 

Hering’s Ready Reference Tables (Conversion Factors).i6mo, morocco, 

Hind’s Inorganic Chemistry.8vo, 

* Laboratory Manual for Students.nmo, 

Holleman’s Text-book of Inorganic Chemistry. (Cooper.).8vo, 

Text-book of Organic Chemistry. (Walker and Mott.).8vo, 

* Laboratory Manual of Organic Chemistry. (Walker.).nmo, 

Hopkins’s Oil-chemists’ Handbook.8vo, 

Jackson’s Directions for Laboratory Work in Physiological Chemistry. .8vo, 

Keep’s Cast Iron.8vo, 

Ladd’s Manual of Quantitative Chemical Analysis.nmo, 

Landauer’s Spectrum Analysis. (Tingle,).8vo, 

* Langworthy and Austen. The Occurrence of Aluminium in Vege able 

Products, Animal Products, and Natural Waters.8vo, 

Lassar-Cohn’s Practical Urinary Analysis. (Lorenz.).i2mo, 

Application of Some General Reactions to Investigations in Organic 

Chemistry. (Tingle.).nmo, 

Leach’s The Inspection and Analysis of Food with Special Reference to State 

Control.8vo, 

Lob’s Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz.).nmo, 
Lodge’s Notes on Assaying and Metallurgical Laboratory Experiments. .. .8vo, 

Lunge’s Techno-chemical Analysis. (Cohn.).nmo, 

Mandel’s Handbook for Bio-chemical Laboratory.nmo, 

* Martin’s Laboratory Guide to Qualitative Analysis with the Blowpipe. . nmo, 
Mason’s Water-supply. (Considered Principally from a Sanitary Standpoint.) 

3d Edition, Rewritten.8vo, 

Examination of Water. (Chemical and Bacteriological.).nmo, 

Matthew’s The Textile Fibres.8vo, 

Meyer’s Determination of Radicles in Carbon Compounds. (Tingle.), .nmo, 

Miller’s Manual of Assaying.nmo, 

Mixter’s Elementary Text-book of Chemistry.nmo, 

Morgan’s Outline of Theory of Solution and its Results..i2ino, 

Elements of Physical Chemistry.nmo, 

Morse’s Calculations used in Cane-sugar Factories.i6mo, morocco, 

Mulliken’s General Method for the Identification of Pure Organic Compounds. 

Vol. I.Large 8vo, 

O’Brine’s Laboratory Guide in Chemical Analysis.8vo, 

O’Driscoll’s Notes on the Treatment of Gold Ores.8vo, 

Ostwald’s Conversations on Chemistry. Part One. (Ramsey.).nmo, 

Ostwald’s Conversations on Chemistry. Part Two. (Turnbull.). (In Press.) 

* Penfield’s Notes on Determinative Mineralogy and Record of Mineral Tests. 

8vo, paper, 

Pictet’s The Alkaloids and their Chemical Constitution. (Biddle.).8vo, 

Pinner’s Introduction to Organic Chemistry. (Austen.).nmo, 

Poole’s Calorific Power of Fuels.8vo, 

Prescott and Winslow’s Elements of Water Bacteriology, with Special Refer¬ 
ence to Sanitary Water Analysis.i2mo, 

4 


1 5C 

2 oe 
5 ot 

3 ok 

12 50 


50 

oc 

oc 

23 

00 

OQ 

50 

5 C 

oc 

75 

50 

50 

00 

00 

25 

50 

00 


3 00 

2 00 
1 00 

I 00 


50 

00 

00 

00 

50 

60 


4 00 
1 25 
3 50 


00 

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50 


5 00 
2 00 
2 00 
1 5 ° 


50 
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I 50 
3 00 

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* Reisig’s Guide to Piece-dyeing. . . -.8vo, 25 co 

Richards and Woodman’s Air, Water, and Food from a Sanitary Standpoint 8vo, 2 00 

Richards’s Cost of Living as Modified by Sanitary Science.i2mo, 1 00 

Cost of Food, a Study in Dietaries.i2mo, 1 00 

* Richards and Williams’s The Dietary Computer.8vo, 1 50 

Ricketts and Russell’s Skeleton Notes upon inorganic Chemistry. (Part I. 

Non-metallic Elements.).8vo, morocco, 75 

Ricketts and Miller’s Notes on Assaying.8vo, 3 00 

Rideal’s Sewage and the Bacterial Purificat'on of Sewage.8vo, 3 50 

Disinfection and the Preservation of Food.8vo, 4 00 

Rigg’s Elementary Manual for the Chemical Laboratory.8vo, 1 25 

Rostoski’s Serum Diagnosis. (Bolduan.).nno, 1 00 

Ruddiman’s Incompatibilities in Prescriptions.8vo, 2 00 

Sabin’s Industrial and Artistic Technology of Paints and Varnish.8vo, 3 00 

Salkowski’s Physiological and Pathological Chemistry. (Orndorff.).8vo, 2 50 

Schimpf’s Text-book of Volumetric Analysis.i2ir.o, 2 50 

Essentials of Volumetric Analysis. . ..nmo, 1 25 

Spencer’s Handbook for Chemists of Beet-sugar Houses.i6mo, mcrccco. 3 00 

Handbook for Sugar Manufacturers and their Chemists. . i6mo, morocco, 2 00 

Stockbridge’s Rocks and Soils.8vo, 2 50 

* Tillman’s Elementary Lessons in Heat.8vo, 1 50 

* Descriptive General Chemistry.8vo, 3 00 

Treadwell’s Qualitative Analysis. (Hall.).8vo. 3 00 

Quantitative Analysis. (Hall.).8vo, 4 00 

Turneaure and Russell’s Public Water-supplies.8vo, 5 00 

Van Deventer’s Physical Chemistry for Beginners. (Boltwood.).nmo, 1 50 

* Walke’s Lectures on Explosives. 8vo, 4 00 

Washington’s Manual of the Chemical Analysis of Rocks.8vo, 2 00 

Wassermann’s Immune Sera: Hsemolyslns, Cytotoxins, and Precipitins. (Bol¬ 
duan.).nmo, r co 

Well’s Laboratory Guide in Qualitative Chemical Analysis.8vo, 1 50 

Short Course in Inorganic Qualitative Chemical Analysis for Engineering 

Students.nmo, 1 50 

Text-book of Chemical Arithmetic. (In press.) 

Whipple’s Microscopy of Drinking-water. 8vo, 3 50 

Wilson’s Cyanide Processes.nmo, r 50 

Chlorination Process.nmo, r 50 

Wulling’s Elementary Course in Inorganic, Pharmaceutical, and Medical 

Chemistry.nmo, 2 00 

CIVIL ENGINEERING. 

BRIDGES AND ROOFS. HYDRAULICS. MATERIALS OF ENGINEERING. 

RAILWAY ENGINEERING. 

Baker’s Engineers’Surveying Instruments.nmo, 3 00 

Bixby’s Graphical Computing Table.Paper 19^X24} inches. 2 5 

** Burr’s Ancient and Modern Engineering and the Isthmian Canal. (Postage, 

27 cents additional.).8vo, 3 50 

Comstock’s Field Astronomy for Engineers.8vo, 2 50 

Davis’s Elevation and Stadia Tables.8vo, 1 00 

Elliott’s Engineering for Land Drainage.nmo, 1 50 

Practical Farm Drainage.nmo, 1 00 

Fiebeger’s Treatise on Civil Engineering. (In press.) 

Folwell’s Sewerage. (DesigningandMaintenar.ee.).8vo, 300 

Freitag’s Architectural Engineering. 2d Edition, Rewritten. 8 vo, 3 5 ° 

French and Ives’s Stereotomy. 8v0 » 2 5 ° 

Goodhue's Municipal Improvements.. 1 75 

Goodrich’s Economic Disposal of Towns’ Refuse. 8v0 » 3 

Gore’s Elements of Geodesy. 8v0 » 2 50 

Hayford’s Text-book of Geodetic Astronomy.evo, 3 00 

Hering’s Ready Reference Tables (Conversion Factors').i6mo, morocco. 2 50 

5 
















































Howe’s Retaining Walls for Earth.i2mo, i 

Johnson’s (J. B.) Theory and Practice of Surveying.Small 8vo, 4 

Johnson’s (L. J.) Statics by Algebraic and Graphic Methods.8vo, 2 

Laplace’s Philosophical Essay on Probabilities. (Truscott and Emory.). i2mo, 2 

Mahan’s Treatise on Civil Engineering. (1873.) (Wood.).8vo, 5 

* Descriptive Geometry.8vo, 1 

Merriman’s Elements of Precise Surveying and Geodesy.8vo, 2 

Elements of Sanitary Engineering.8vo, 2 

Merriman and Brooks’s Handbook for Surveyors.i6mo, morocco, 2 

Nugent’s Plane Surveying.8vo, 3 

Ogden’s Sewer Design.nmo, 2 

Patton’s Treatise on Civil Engineering.8vo half leather, 7 

Reed’s Topographical Drawing and Sketching. 4to, 5 

Rideal’s Sewage and the Bacterial Purification of Sewage.8vo, 3 

Siebert and Biggin’s Modern Stone-cutting and Masonry.8vo, 1 

Smith’s Manual of Topographical Drawing. (McMillan.).8vo, 2 

Sondericker’s Graphic Statics, with Applications to Trusses, Beams, and Arches. 

8 vo, 2 

Taylor and Thompson’s Treatise on Concrete, Plain and Reinforced.8vo, 5 

* Trautwine’s Civil Engineer’s Pocket-book.i6mo, morocco, 5 

Wait’s Engineering and Architectural Jurisprudence.8vo, 6 

Sheep, 6 

Law of Operations Preliminary to Construction in Engineering and Archi¬ 
tecture. 8vo, 5 

Sheep, 5 

Law of Contracts.8vo, 3 

Warren’s Stereotomy—Problems in Stone-cutting.8vo, 2 

Webb’s Problems in the Use and Adjustment of Engineering Instruments. 

i6mo, morocco, 1 

* Wheeler s Elementary Course of Civil Engineering.8vo, 4 

Wilson’s Topographic Surveying.8vo, 3 

BRIDGES AND ROOFS. 

Boiler’s Practical Treatise on the Construction of Iron Highway Bridges. . 8vo, 2 

* Thames River Bridge.4to, paper, 5 

Burr’s Course on the Stresses in Bridges and Roof Trusses, Arched Ribs, and 

Suspension Bridges..:.. ..8vo, 3 

Burr and Falk’s Influence Lines for Bridge and Roof Computations . . . 8vo, 3 

Du Bois’s Mechanics of Engineering. Vol. II.Small 4to, 10 

Foster’s Treatise on Wooden Trestle Bridges.4to, 5 

Fowler’s Ordinary Foundations.8vo, 3 

Greene’s Roof Trusses.8vo, 1 

Bridge Trusses.8vo, 2 

Arches in Wood, Iron, and Stone.8vo, 2 

Howe’s Treatise on Arches.8vo, 4 

Design of Simple Roof-trusses in Wood and Steel.8vo, 2 

Johnson, Bryan, and Turneaure’s Theory and Practice in the Designing of 

Modern Framed Structures..Small 410, 10 

Merriman and Jacoby’s Text-book on Roofs and Bridges: 

Part I. Stresses in Simple Trusses.8vo, 2 

Part II. Graphic Statics. 8vo, 2 

Part III. Bridge Design...8vo, 2 

PartIV. Higher Structures.8vo, 2 

Morison’s Memphis Bridge.4to, 10 

Waddell’s De Pontibus, a Pocket-book for Bridge Engineers. . i6mo, morocco, 3 

Specifications for Steel Bridges.. . nmo, 1 

Wood’s Treatise on the Theory of the Construction of Bridges and Roofs. .8vo, 2 
Wright’s Designing of Draw-spans: 

Part I. Plate-girder Draws.8vo, 2 

Part II. Riveted-truss and Pin-connected Long-span Draws.8vo, 2 

Two parts in one volume.,..8vo, 3 


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HYDRALLICS 


Bazin’s Experiments upon the Contraction of the Liquid Vein Issuing from 

an Orifice. (Trautwine.). 8vo, 2 00 

Bovey’s Treatise on Hydraulics. .8vo, 5 00 

Church’s Mechanics of Engineering.8vo, 6 00 

Diagrams of Mean Velocity of Water in Open Channels.payer, 1 50 

Coffin’s Graphical Solution of Hydraulic Problems.i6mo, morocco, 2 50 

Flather’s Dynamometers, and the Measurement of Power.nmo, 3 00 

Folwell’s Water-supply Engineering.8vo, 4 00 

Frizell’s Water-power.8vo, 5 00 

Fuertes’s Water and Public Health.nmo, 1 50 

Water-filtration Works.i2mo, 2 50 

Ganguillet and Kutter’s General Formula for the Uniform Flow of Water in 

Rivers and Other Channels. (Hering and Trau vine.)...8vo 400 

Hazen’s Filtration of Public Water-supply.8vo, 3 00 

Hazlehurst’s Towers and Tanks for Water-works.8vo, 2 50 

Herschel’s 115 Experiments on the Carrying Capacity of Large, Riveted, Metal 

Conduits.8vo, 2 00 

Mason’s Water-supply. (Considered Principally from a Sanitary Standpoint.) 

8vo, 4 00 

Merriman’s Treatise on Hydraulics.8vo, 5 00 

* Michie’s Elements of Analytical Mechanics.. . .8vo, 4 00 

Schuyler’s Reservoirs for Irrigation, Water-power, and Domestic Water- 

supply.Large 8vo, 5 00 

** Thomas and Watt’s Improvement of Rivers. (Post., 44c. additional.).4to, 6 00 

Turneaure and Russell’s Public Water-supplies.8vo, 5 00 

Wegmann’s Design and Construction of Dams.4to, 5 00 

Water-supply of the City of New York from 1658 to 1895.4to, 10 00 

Wilson’s Irrigation Engineering.Small 8vo, 4 00 

Wolff’s Windmill as a Prime Mover. 8vo, 3 00 

Wood’s Turbines.%.8vo, 2 50 

Elements of Analytical Mechanics. 8vo, 3 00 

MATERIALS OF ENGINEERING. 

Baker’s Treatise on Masonry Construction... < . .8vo, 5 00 

Roads and Pavements. 3 vo, 5 00 

Black’s United States Public Works.Oblong 4to> 5 00 

Bovey’s Strength of Materials and Theory of Structures.8vo, 7 50 

Burr’s Elasticity and Resistance of the Materials of Engineering.8vo, 7 50 

Byrne’s Highway Construction. . ..8vo, 5 00 

Inspection of the Materials and Workmanship Employed in Construction. 

i6mo, 3 00 

Church’s Mechanics of Engineering.8vo, 6 00 

Du Bois’s Mechanics of Engineering. Vol. I.Small 4to, 7 SO 

Johnson’s Materials of Construction.Large 8vo, 6 00 

Fowler’s Ordinary Foundations.8vo, 3 50 

Keep's Cast Iron.8vo, 2 50 

Lanza’s Applied Mechanics. 8 vo, 7 5 ° 

Marten’s Handbook on Testing Materials. (Henning.) 2 vols.8vo, 7 50 

Merrill’s Stones for Building and Decoration. 8vo, 5 00 

Merriman’s Text-book on the Mechanics of Materials.8vo, 4 00 

Strength of Materials.. 1 00 

Metcalf’s Steel. A Manual for Steel-users.i2mo, 2 00 

Patton’s Practical Treatise on Foundations.8vo, 5 00 

Richardson’s Modern Asphalt Pavements.8vo, 3 00 

Richey’s Handbook for Superintendents of Construction.ibmo, mor., 4 00 

Rockwell’s Roads and Pavements in France.i 2 mo, 1 25 

7 















































Sabin’s Industrial and Artistic Technology of Paints and Varnish.8vo, 

Smith’s Materials of Machines.i2mo, 

Snow’s Principal Species of Wood.8vo, 

Spalding’s Hydraulic Cement.i2mo, 

Text-book on Roads and Pavements.i2mo f 

Taylor and Thompson’s Treatise on Concrete, Plain and Reinforced.8vo, 

Thurston’s Materials of Engineering. 3 Parts.8vo, 

Part I. Non-metallic Materials of Engineering and Metallurgy.8vo, 

Part II. Iron and Steel.8vo, 

Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents.8vo, 

Thurston’s Text-book of the Materials of Construction.8vo, 

Tillson’s Street Pavements and Paving Materials.8vo, 

Waddell’s De Pontibus. ( A Pocket-book for Bridge Engineers.). .i6mo, mor., 

Specifications for Stc. 1 Bridges.nmo, 

Wood’s (De V.) Treatise on the Resistance of Materials, and an Appendix on 

the Preservation of Timber. 8vo, 

Wood’s (De V.) Elements of Analytical Mechanics.8vo, 

Wood’s (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and 
Steel.8vo, 


RAILWAY ENGINEERING. 


Andrew’s Handbook for Street Railway Engineers.3x5 inches, morocco, 

Berg’s Buildings and Structures of American Railroads.4to, 

Brook’s Handbook of Street Railroad Location.i6mo, morocco, 

Butt’s Civil Engineer’s Field-book.i6mo, morocco, 

Crandall’s Transition Curve.i6mo, morocco, 

Railway and Other Earthwork Tables.8vo, 

Dawson’s “Engineering” and Electric Traction Pocket-book. .i6mo, morocco, 
Dredge’s History of the Pennsylvania Railroad: (1879).Paper, 

* Drinker’s Tunnelling, Explosive Compounds, and Rock Drills.4to, half mor., 

Fisher’s Table of Cubic Yards.Cardboard, 

Godwin’s Railroad Engineers’ Field-book and Explorers’ Guide. . . i6mo, mor., 

Howard’s Transition Curve Field-book.i6mo, morocco, 

Hudson’s Tables for Calculating the Cubic Contents of Excavations and Em¬ 
bankments.8vo, 

Molitor and Beard’s Manual for Resident Engineers.i6mo, 

Nagle’s Field Manual for Railroad Engineers.i6mo, morocco, 

Philbrick’s Field Manual for Engineers.i6mo, morocco, 

Searles’s Field Engineering.i6mo, morocco, 

Railroad Spiral.i6mo, morocco, 

Taylor's Prismoidal Formulae and Earthwork.8vo, 

* Trautwine’s Method of Calculating the Cube Contents of Excavations and 

Embankments by the Aid of Diagrams.8vo, 

The Field Practice of Laying Out Circular Curves for Railroads. 

nmo, morocco, 

Cross-section Sheet.Paper, 

Webb’s Railroad Construction.i6mo, morocco, 

Wellington’s Economic Theory of the Location of Railways.Small 8vo, 


DRAWING. 

Barr’s Kinematics of Machinery.8vo, 

* Bartlett’s Mechanical Drawing.8vo, 

* “ “ “ Abridged Ed.8vo, 

Coolidge’s Manual of Drawing.8vo, paper 

Coolidge and Freeman’s Elements of General Drafting for Mechanical Engi¬ 
neers.Oblong 4to, 

Durley’s Kinematics of Machines.8vo, 

Emch’s Introduction to Projective Geometry and its Applications.8vo. 

8 


3 00 

1 00 
3 50 

2 00 
2 00 
5 00 
8 00 

2 00 

3 50 

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5 00 

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1 25 

2 00 

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1 25 
5 00 

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Hill’s Text-book on Shades and Shadows, and Perspective.8vo, 2 00 

Jamison’s Elements of Mechanical Drawing.8vo, 2 50 

Jones’s Machine Design: 

Parti. Kinematics of Machinery.8vo, 1 50 

Part II. Form, Strength, and Proportions of Parts.8vo, 300 

MacCord’s Elements of Descriptive Geometry.8vo, 3 00 

Kinematics; or, Practical Mechanism.8vo, 5 00 

Mechanical Drawing.4to, 4 00 

Velocity Diagrams.8vo, 1 50 

* Mahan’s Descriptive Geometry and Stone-cutting..8vo, 1 50 

Industrial Drawing. (Thompson.).8vo, 3 50 

Moyer’s Descriptive Geometry.8vo, 2 00 

Reed’s Topographical Drawing and Sketching.4to, 5 00 

Reid’s Course in Mechanical Drawing. 8vo, 2 00 

Text-book of Mechanical Drawing and Elementary Machine Design.8vo, 3 00 

Robinson’s Principles of Mechanism. . . . !.8vo, 3 00 

Schwamb and Merrill’s Elements of Mechanism.8vo, 3 00 

Smith’s Manual of Topographical Drawing. (McMillan.).8vo, 2 50 

Warren’s Elements of Plane and Solid Free-hand Geometrical Drawing. i2mo, 1 00 

Drafting Instruments and Operations.i2mo s 1 25 

Manual of Elementary Projection Drawing.nmo, 1 50 

Manual of Elementary Problems in the Linear Perspective of Form and 

Shadow. nmo, 1 00 

Plane Problems in Elementary Geometry.nmo, 1 25 

Primary Geometry.nmo, 75 

Elements of Descriptive Geometry, Shadows, and Perspective.8vo, 3 50 

General Problems of Shades and Shadows.8vo, 3 00 

Elements of Machine Construction and Drawing.8vo, 7 50 

Problems, Theorems, and Examples in Descriptive Geometry.8vo, 2 50 

Weisbach’s Kinematics and Power of Transmission. (Hermann and Klein)8vo, 5 00 

Whelpley’s Practical Instruction in the Art of Letter Engraving.nmo, 2 00 

Wilson’s (H. M.) Topographic Surveying.8vo, 3 50 

Wilson’s (V. T.) Free-hand Perspective.8vo, 2 50 

Wilson’s (V. T.) Free-hand Lettering.8vo, x 00 

Woolf’s Elementary Course in Descriptive Geometry.Large 8vo, 3 00 

ELECTRICITY AND PHYSICS. 

Anthony and Brackett’s Text-book of Physics. (Magie.).Small 8vo, 3 00 

Anthony’s Lecture-notes on the Theory of Electrical Measurements.. . .i2mo, 1 00 

Benjamin’s History of Electricity.8vo, 3 00 

Voltaic Cell.8vo, 3 00 

Classen’s Quantitative Chemical Analysis by Electrolysis. (Boltwood.).8vo, 3 00 

Crehore and Squier’s Polarizing Photo-chronograph.8vo, 3 00 

Dawson’s “Engineering” and Electric Traction Pocket-book. i6mo, morocco, 5 00 
Dolezalek’s Theory of the Lead Accumulator (Storage Battery). (Von 

Ende.).i2mo, 2 50 

Duhem’s Thermodynamics and Chemistry. (Burgess.).8vo, 4 00 

Flather’s Dynamometers, and the Measurement of Power.i2mo, 3 00 

Gilbert’s De Magnete. (Mottelay.).8vo, 250 

Hanchett’s Alternating Currents Explained.i2mo, 1 00 

Hering’s Ready Reference Tables (Conversion Factors).i6mo, morocco, 2 50 

Holman’s Precision of Measurements.8vo, 2 00 

Telescopic Mirror-scale Method, Adjustments, and Tests-Large 8vo, 75 

Kinzbrunner's Testing of Continuous-Current Machines.8vo t 2 00 

Landauer’s Spectrum Analysis. (Tingle.).8vo, 3 00 

Le Chatelien s High-temperature Measurements. (Boudouard— Burgess.) i2mo, 3 00 
Lob’s Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz.) i2mo, 1 00 

9 













































4 Lyons’s Treatise on Electromagnetic Phenomena. Vols. I. and II. 8vo, each, 6 oo 


* Michie’s Elements of Wave Motion Relating to Sound and Light. .8vo, 4 00 

Niaudet’s Elementary Treatise on Electric Batteries. (Fishback.).nmo, 2 50 

* Rosenberg’s Electrical Engineering. (Haldane Gee—Kinzbrunner.). . .8vo, 1 50 

Ryan, Norris, and Hoxie’s Electrical Machinery. Vol. 1.8vo, 2 50 

Thurston’s Stationary Steam-engines.8vo, 2 50 

* Tillman’s Elementary Lessons in Heat.8vo, 1 50 

Tory and Pitcher’s Manual of Laboratory Physics.Small 8vo, 2 00 

Ulke’s Modern Electrolytic Copper Refining.*.8vo, 3 00 


LAW. 


* Davis’s Elements of Law.8vo, 2 50 

* Treatise on the Military Law of United States.8vo, 7 00 

* Sheep, 7 50 

Manual for Courts-martial.i6mo, morocco, 1 50 

Wait’s Engineering and Architectural Jurisprudence.8vo, 6 00 

Sheep, 6 50 

Law of Operations Preliminary to Construction in Engineering and Archi¬ 
tecture.8vo, 5 00 

Sheep, 5 50 

Law of Contracts.8vo, 3 00 

Winthrop’s Abridgment of Military Law.nmo, 2 50 


MANUFACTURES. 


. Bernadou’s Smokeless Powder—Nitro-cellulose and Theory of the Cellulose 

Molecule.nmo, 2 50 

Bolland’s Iron Founder..12mo, 2 50 

“ The Iron Founder,” Supplement.nmo, 2 50 

Encyclopedia of Founding and Dictionary of Foundry Terms Used in the 

Practice of Moulding.nmo, 3 00 

Eissler’s Modern High Explosives.8vo, 4 00 

Effront’s Enzymes and their Applications. (Prescott.).8vo, 3 00 

Fitzgerald’s Boston Machinist.nmo, 1 00 

Ford’s Boiler Making for Boiler Makers.. i8mo, 1 00 

Hopkin’s Oil-chemists’ Handbook.8vo, 3 00 

Keep’s Cast Iron.8vo, 2 50 

Leach’s The Inspection and Analysis of Food with Spec.al Reference to State 

Control.Large 8vo, 7 50 

Matthews’s The Textile Fibres.8vo, 3 50 

Metcalf’s Steel. A Manual for Steel-users.nmo, 2 00 

Metcalfe’s Cost of Manufactures—And the Administration of Workshops.8vo, 5 00 

Meyer’s Modern Locomotive Construction.4to, 10 00 

Morse’s Calculations used in Cane-sugar Factories.i6mo, morocco, 1 50 

* Reisig’s Guide to Piece-dyeing.8vo, 25 00 

Sabin’s Industrial and Artistic Technology of Paints and Varnish.8vo, 3 00 

Smith’s Press-working of Metals.8vo, 3 00 

Spalding’s Hydraulic Cement.nmo, 2 00 

Spencer’s Handbook for Chemists of Beet-sugar Houses. ... i6mo, morocco, 3 00 

Handbook for Sugar Manufacturers and their Chemists. . i6mo, morocco, 2 00 

Taylor and Thompson’s Treatise on Concrete, Plain and Reinforced.8vo, 5 00 

Thurston’s Manual of Steam-boilers, their Designs, Construction and Opera¬ 
tion.8vo, 5 00 

* Walke’s Lectures on Explosives.8vo, 4 00 

Ware’s Manufacture of Sugar. (In press.) 

West’s American Foundry Practice. nmo, 2 50 

Moulder’s Text-book.12mo, 2 50 


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Wolff’s Windmill as a Prime Mover.g vo> 

Wood’s Rustless Coatings: Corrosion and Electrolysis of Iron and Steel. .8vo, 


MATHEMATICS. 


Baker’s Elliptic Functions..8vo, 

* Bass’s Elements of Differential Calculus.i2mo, 

Briggs’s Elements of Plane Analytic Geometry.i2mo, 

Compton’s Manual of Logarithmic Computations.i2mo, 

Davis’s Introduction to the Logic of Algebra.8vo, 

* Dickson’s College Algebra.Large nmo, 

* Introduction to the Theory of Algebraic Equations.Large nmo, 

Emch’s Introduction to Projective Geometry and its Applications.8vo, 

Halsted’s Elements of Geometry.•.8vo, 

Elementary Synthetic Geometry.8vo, 

Rational Geometry.nmo, 

* Johnson’s (J. B.) Three-place Logarithmic Tables: Vest-pocket size.paper, 

xoo copies for 

* Mounted on heavy cardboard, 8X10 inches, 

io copies for 

Johnson’s (W. W.) Elementary Treatise on Differential Calculus. .Small 8vo, 
Johnson’s (W. W.) Elementary Treatise on the Integral Calculus. Small 8vo, 

Johnson’s (W. W.) Curve Tracing in Cartesian Co-ordinates.nmo, 

Johnson’s (W. W.) Treatise on Ordinary and Partial Differential Equations. 

Small 8vo, 

Johnson’s (W. W.) Theory of Errors and the Method of Least Squares, nmo, 

* Johnson’s (W. W.) Theoretical Mechanics.nmo, 

Laplace’s Philosophical Essay on Probabilities. (Truscott and Emory.). i2mo, 

Ludlow and Bass’. Elements of Trigonometry and Logarithmic and Other 

Tables. 8vo, 

Trigonometry and Tables published separately.Each, 

* Ludlow’s Logarithmic and Trigonometric Tables.8vo, 

Maurer’s Technical Mechanics. 8.., 

Merriman and Woodward’s Higher Mathematics.8vo, 

Merriman’s Method of Least Squares.8vo, 

Rice and Johnson’s Elementary Treatise on the Differential Calculus.. Sm. 8vo, 

Differential and Integral Calculus. 2 vols. in one.Small 8vo, 

Wood's Elements of Co-ordinate Geometry.8vo, 

Trigonometry: Analytical, Plane, and Spherical.nmo, 


MECHANICAL ENGINEERING. 

MATERIALS OF ENGINEERING, STEAM-ENGINES AND BOILERS. 


Bacon’s Forge Practice.nmo, 

Baldwin’s Steam Heating for Buildings.nmo, 

Barr’s Kinematics of Machinery.8vo, 

* Bartlett’s Mechanical Drawing.8vo, 

<* “ “ “ Abridged Ed.8vo, 

Benjamin’s Wrinkles and Recipes.nmo, 

Carpenter’s Experimental Engineering. 8vo, 

Heating and Ventilating Buildings.8vo, 

Cary’s Smoke Suppression in Plants using Bituminous Coal. (In Prepara¬ 
tion.) 

Clerk’s Gas and Oil Engine.Small 8vo, 

Coolidge’s Manual of Drawing. .8vo, paper, 

Coolidge and Freeman’s Elements of General Drafting for Mechanical En¬ 
gineers...Oblong 4to, 


3 00 

4 00 


x 50 

4 00 
1 #00 
i 50 
i 50 
1 50 

1 25 

2 50 
i 75 
1 50 

1 75 
1 5 

5 00 

25 

2 00 
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1 50 
1 00 

3 50 

1 50 
3 00 

2 00 

3 00 
2 00 

1 00 

4 00 

5 00 

2 00 

3 00 
2 50 
2 00 
1 00 


1 50 

2 50 
2 50 

3 OO 

1 50 

2 OO 

6 00 

4 00 

4 00 

1 OO 

2 50 


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Cromwell’s Treatise on Toothed Gearing.i2mo, i 50 

Treatise on Belts and Pulleys.i2mo, 1 50 

Durley’s Kinematics of Machines.8vo, 4 00 

Flather’s Dynamometers and the Measurement of Power.nmo, 3 00 

Rope Driving.nmo, 2 00 

Gill’s Gas and Fuel Analysis for Engineers.nmo, 1 25 

Hall’s Car Lubrication.1.nmo, 1 00 

Hering’s Ready Reference Tables (Conversion Factors).i6mo, morocco, 2 50 

Hutton’s The Gas Engine.8vo, 5 00 

Jamison’s Mechanical Drawing. 8vo, 2 50 

Jones’s Machine Design: 

Parti. Kinematics of Machinery.8vo, 1 50 

Part II. Form, Strength, and Proportions of Parts.8vo, 3 00 

Kent’s Mechanical Engineers’ Pocket-book.i6mo, morocco, 5 00 

Kerr’s Power and Power Transmission.8vo, 2 00 

Leonard’s Machine Shop, Tools, and Methods. (In press.) 

Lorenz’s Modern Refrigerating Machinery. (Pope, Haven, and Dean.) (In press.) 

MacCord’s Kinematics; or, Practical Mechanism.8vo, 5 00 

Mechanical Drawing.4to, 4 00 

Velocity Diagrams.*.8vo, 1 50 

Mahan’s Industrial Drawing. (Thompson.).8vo, 3 50 

Poole's Calorific Power of Fuels.8vo, 3 00 

Reid’s Course in Mechanical Drawing.8vo, 2 00 

Text-book of Mechanical Drawing and Elementary Machine Design.8vo, 3 00 

Richard’s Compressed Air.i2mo, 1 50 

Robinson’s Principles of Mechanism.8vo, 3 00 

Schwamb and Merrill’s Elements of Mechanism.8vo, 3 00 

Smith’s Press-working of Metals.8vo, 3 00 

Thurston’s Treatise on Friction and Lost Work in Machinery and Mill 

Work.8vo, 3 00 

Animal as a Machine and Prime Motor, and the Laws of Energetics. i2mo, 1 00 

Warren’s Elements of Machine Construction and Drawing. .8vo, 7 50 

Weisbach’s Kinematics and the Power of Transmission. (Herrmann— 

Klein.).8vo,. 5 00 

Machinery of Transmission and Governors. (Herrmann—Klein.). .8vo, 5 00 

Wolff’s Windmill as a Prime Mover.8vo, 3 00 

Wood’s Turbines.8vo, 2 50 

MATERIALS OF ENGINEERING. 

Bovey’s Strength of Materials and Theory of Structures.8vo, 7 50 

Burr’s Elasticity and Resistance ot the Materials of Engineering. 6th Edition. 

Reset.8vo, 7 50 

Church’s Mechanics of Engineering.8vo, 6 00 

Johnson’s Materials of Construction.8vo, 6 00 

Keep’s Cast Iron.8vo, 2 50 

Lanza’s Applied Mechanics.8vo, 7 50 

Martens’s Handbook on Testing Materials. (Henning.).8vo, 7 50 

Merriman’s Text-book on the Mechanics of Materials.8vo, 4 00 

Strength of Materials.i2mo, 1 00 

Metcalf’s Steel. A manual for Steel-users.nmo. 2 00 

Sabin’s Industrial and Artistic Technology of Paints and Varnish.8vo, 3 00 

Smith’s Materials of Machines.nmo, 1 00 

Thurston’s Materials of Engineering.3 vols., 8vo, 8 00 

Part II. Iron and Steel.8vo, 3 50 

Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents.8vo, 2 50 

Text-book of the Materials of Construction.8vo, 5 o* 

12 
















































Wood’s (De V.) Treatise on the Resistance of Materials and an Appendix on 

the Preservation of Timber.8vo, 2 oo 

Wood’s (De V.) Elements of Analytical Mechanics.8vo, 3 00 

food’s (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and 

Steel... 4 00 

STEAM-ENGINES AND BOILERS. 

Berry’s Temperature-entropy Diagram.i2mo, 1 25 

Carnot’s Reflections on the Motive Power of Heat. (Thurston.).i2mo, 1 50 

Dawson’s “Engineering” and Electric Traction Pocket-book... .i6mo, mor., 5 00 

Ford’s Boiler Making for Boiler Makers.i8mo, 1 00 

Goss’s Locomotive Sparks.8vo, 2 00 

Hemenway’s Indicator Practice and Steam-engine Economy.nmo, 2 00 

Hutton’s Mechanical Engineering of Power Plants.8vo, 5 00 

Heat and Heat-engines.8vo, 5 00 

Kent’s Steam boiler Economy.8vo, 4 00 

Kneass’s Practice and Theory of the Injector.8vo, 1 50 

MacCord’s Slide-valves.8vo, 2 00 

Meyer’s Modern Locomotive Construction.4to, 10 00 

Peabody’s Manual of the Steam-engine Indicator.i2mo, 1 50 

Tables of the Properties of Saturated Steam and Other Vapors.8vo, 1 00 

Thermodynamics of the Steam-engine and Other Heat-engines.8vo, 5 00 

Valve-gears for Steam-engines.8vo, 2 50 

Peabody and Miller’s Steam-boilers.8vo, 4 00 

Pray’s Twenty Years with the Indicator.Large 8vo, 2 50 

Pupin’s Thermodynamics of Reversible Cycles in Gases and Saturated Vapors. 

(Osterberg.).i2mo, 1 25 

Reagan’s Locomotives: Simple Compound, and Electric.i2mo, 2 50 

Rontgen’s Principles of Thermodynamics. (Du Bois.).8vo, 5 00 

Sinclair’s Locomotive Engine Running and Management.i2mo, 2 00 

Smart’s Handbook of Engineering Laboratory Practice.i2mo, 2 50 

%iow’s Steam-boiler Practice.8vo, 3 00 

Spangler’s Valve-gears.8vo, 2 50 

Notes on Thermodynamics.i2mo, 1 00 

Spangler, Greene, and Marshall’s Elements of Steam-engineering.8vo, 3 00 

Thurston’s Handy Tables.8vo, 1 50 

Manual of the Steam-engine.2 vols., 8vo, xo 00 

Part I. History, Structure, and Theory.: .8vo, 6 00 

Part H. Design, Construction, and Operation.8vo, 6 00 

Handbook of Engine and Boiler Trials, and the Use of the Indicator and 

the Prony Brake.8vo, 5 00 

Stationary Steam-engines.8vo, 2 50 

Steam-boiler Explosions in Theory and in Practice.i2mo, 1 50 

Manual of Steam-boilers, their Designs, Construction, and Operation.8vo, 5 00 

Weisbach’s Heat. Steam, and Steam-engines. (Du Bois.).8vo, 5 00 

Whitham’s Steam-engine Design.8vo, 5 00 

Wilson’s Treatise on Steam-boilers. (Flather.).i6mo, 2 50 

Wood’s Thermodynamics, Heat Motors, and Refrigerating Machines.. .8vo, 4 00 

MECHANICS AND MACHINERY. 

Barr’s Kinematics of Machinery.8vo, 2 50 

Bovey’s Strength of Materials and Theory of Structures.8vo, 7 50 

Chase’s The Art of Pattern-making.l 2 mo, 2 50 

Church’s Mechanics of Engineering.8vo, 6 00 


13 













































Church's Notes and Examples in Mechanics.8vo, 2 00 

Compton’s First Lessons in Metal-working.i2mo, 1 50 

Compton and De Groodt’s The Speed Lathe.nmo, 1 50 

Cromwell’s Treatise on Toothed Gearing.nmo, 1 50 

Treatise on Belts and Pulleys. nmo, 1 50 

Dana’s Text-book of Elementary Mechanics for Colleges and Schools, .nmo, 1 50 

Dingey’s Machinery Pattern Making.nmo, 2 00 

Dredge’s Record of the Transportation Exhibits Building of the World’s 

Columbian Exposition of 1893.4to half morocco, 5 00 

Du Bois’s Elementary Principles of Mechanics: 

Vol. I. Kinematics.8vo, 3 50 

Vol. II. Statics.8vo, 4 00 

Vol. III. Kinetics.8vo, 3 50 

Mechanics of Engineering. Vol. I.Small 4to, 7 50 

VoL II.Small 4to, 10 00 

Durley’s Kinematics of Machines.8vo, 4 00 

Fitzgerald’s Boston Machinist. .. .i6mo, 1 00 

Flather’s Dynamometers, and the Measurement of Power.nmo, 3 00 

Rope Driving.nmo, 2 00 

Goss’s Locomotive Sparks.8vo, 2 00 

Hall’s Car Lubrication..12mo, 1 00 

Holly’s Art of Saw Filing.i8mo, 75 

James’s Kinematics of a Point and the Rational Mechanics of a Particle. Sm.8ve,2 00 

* Johnson’s (W. W.) Theoretical Mechanics.nmo, 3 00 

Johnson’s (L. J.) Statics by Graphic and Algebraic Methods.8vo, 2 00 

Jones’s Machine Design: 

Part I. Kinematics of Machinery.8vo, 1 50 

Part II. Form, Strength, and Proportions of Parts.8vo, 3 00 

Kerr’s Power and Power Transmission. 8vo, 2 00 

Lanza’s Applied Mechanics. ..8vo, 7 50 

Leonard’s Machine Shop, Tools, and Methods. (In press.) 

Lorenz’s Modern Refrigerating Machinery. (Pope, Haven, and Dean.) (In press.) 

MacCord’s Kinematics; or, Practical Mechanism.8vo, 5 00 

Velocity Diagrams.8vo, 1 50 

Maurer’s Technical Mechanics.8vo, 4 00 

Merriman’s Text-book on the Mechanics of Materials.. . 8vo, 4 00 

* Elements of Mechanics.nmo, 1 00 

* Michie’s Elements of Analytical Mechanics.8vo, 4 00 

Reagan’s Locomotives: Simple, Compound, and Electric.nmo, 2 50 

Reid’s Course in Mechanical Drawing.8vo, 2 00 

Text-book of Mechanical Drawing and Elementary Machine Design . 8vo, 3 00 

Richards’s Compressed Air.nmo, 1 50 

Robinson’s Principles of Mechanism.8vo, 3 00 

Ryan, Norris, and Hoxie’s Electrical Machinery. Vol. 1 .8vo, 2 50 

Schwamb and Merrill’s Elements of Mechanism.8vo, 3 00 

Sinclair’s Locomotive-engine Running and Management.*..nmo, 2 00 

Smith’s (O.) Press-working of Metals.8vo, 3 00 

Smith’s (A. W.) Materials of Machines.nmo, 1 00 

Spangler, Greene, and Marshall’s Elements of Steam-engineering.8vo, 3 00 

Thurston’s Treatise on Friction and Lost Y/ork in Machinery and Mill 

Work.8vo, 3 00 

Animal as a Machine and Prime Motor, and the Laws of Energetics. 

nmo, 1 00 

Warren’s Elements of Machine Construction and Drawing.8vo, 7 50 

Weisbach’s Kinematics and Power of Transmission. (Herrmann—Klein. ),8vo, 5 00 

Machinery of Transmission and Governors. (Herrmann—Klein.).8vo, 5 00 

Wood’s Elements of Analytical Mechanics.8vo, 3 00 

Principles of Elementary Mechanics.i2mo, 1 25 

Turbines.8vo. 2 50 

The World’s Columbian Exposition of 1893.4to, 1 00 

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METALLURGY. 


Fgleston’s Metallurgy of Silver, Gold, and Mercury: 

Vol. I. Silver...8vo, 7 50 

Vol. II. Gold and Mercury.8vo, 7 50 

** Iles’s Lead-smelting. (Postage 9 cents additional).i2mo, 2 50 

Keep’s Cast Iron..*.8vo, 2 50 

Kunhardt’s Practice of Ore Dressing in Europe.8vo, 1 50 

Le Chatelier’s High-temperature Measurements. (Boudouard—Burgess.)i2mo, 3 00 

Metcalf’s Steel. A Manual for Steel-users- ..i2mo, 2 00 

, Smith’s Materials of Machines.i2mo, 1 00 

Thurston’s Materials of Engineering. In Three Parts.8vo, 8 00 

Part II. Iron and Steel.8vo, 3 50 

Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents.8vo, 2 50 

Ulke’s Modern Electrolytic Copper Refining.8vo, 3 00 

MINERALOGY. 


Barringer’s Description of Minerals of Commercial Value. Oblong, morocco, 

Boyd’s Resources of Southwest Virginia.8vo 

Map of Southwest Virignia...Pocket-book form. 

Brush’s Manual of Determinative Mineralogy. (Penfirdd.).8vo, 

Chester’s Catalogue of Minerals. 8vo, paper, 

Cloth, 

Dictionary of the Names of Minerals.. .8vo, 

Dana’s System of Mineralogy.Large 8vo, half leather, 12 

First Appendix to Dana’s New “ System of Mineralogy.”.Large 8vo, 1 

Text-book of Mineralogy..8vo, 

Minerals and How to Study Them.nmo, 

Catalogue of American Localities of Minerals.Large 8vo, 

Manual of Mineralogy and Petrography.nmo. 

Douglas’s Untechnical Addresses on Technical Subjects.nmo, 

Eakle’s Mineral Tables.8vo, 

Egleston’s Catalogue of Minerals and Synonyms.8vo, 

Hussak’s The Determination of Rock-forming Minerals. (Smith.) .Small 8vo, 
Merrill’s Non-metallic Minerals; Their Occurrence and Uses.8vo, 

* P&nfield’s Notes on Determinative Mineralogy and Record of Mineral Tests. 

8vo paper, o 

Roseubusch’s Microscopical Physiography of the Rock-making Minerals. 

(Iddings.). ..8vo, 5 

* Tillman's Text-book of Important Minerals and Rocks. 8vo, 2 

Willi&ms’s Manual of Lithology. 8vo, 3 


MINING. 

beard’s Ventilation of Mines...i2mo. 2 

Boyd’s Resources of Southwest Virginia.8vo, 3 

M*,p of Southwest Virginia.Pocket book form, 2 

Douglar.'s Untechnical Addresses on Technical Subjects.i2mo, 1 

* Drirdier’s Tunneling, Explosive Compounds, and Rock Drills. ,4to,hf. mor.. 25 

Eissler'o Modern High Explosives.. . 8vo. 4 

Fowler’s Sewage Works Analyses.nmo 2 

Goodyear’s Coal-mines of the Western Coast of the United States.. .. .nmo, 2 

Ihlseng’s Manual of Mining. . .8vo, 5 

** Iles’s Lead-smelting. (Postage pc. additional.)...nmo, 2 

Kunhardt’s Practice of Ore Dressing in Europe.8vo, 1 

O’Driscoll’s Notes on the Treatment of Gold Ores...8vo, 2 

* Walke's Lectures on Explosives....8vo, 4 

Wilson’s Cyanide Processes.nmo, 1 

Chic*"nation Process...nmo, x 


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50 

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Wilson’s HydrauL,. and Placer Mining.i2mo, 2 09 

Treatise on Practical and Theoretical Mine Ventilation.nmo, 1 25 

SANITARY SCIENCE. 

Folwell’s Sewerage. (Designing, Construction, and Maintenance.).8vo, 3 oc 

Water-supply Engineering. 8vo, 4 oo 

Fuertes’s Water and Public Health.i2mo, 1 50 

Water-filtration Works. nmo, 2 50 

Gerhard’s Guide to Sanitary House-inspection.i6mo, 1 00 

Goodrich’s Economic Disposal of Town’s Refuse.Demy 8vo, 3 50 

Hazen’s Filtration of Public Water-supplies.8vo, 3 00 

Leach’s The Inspection and Analysis of Food with Special Reference to State 

Control.8vo, 7 50 

Mason’s Water-supply. (Considered principally from a Sanitary Staudpoint) 8vo, 4 00 

Examination of Water. (Chemical and Bacteriological.).nmo, 1 25 

Merriman’s Elements of Sanitary Eng/neering.8vo, 2 00 

Ogden’s Sewer Design.nmo,- 2 00 

Prescott and Winslow’s Elements of Water Bacteriology, with Special Refer¬ 
ence to Sanitary Water Analysis.nmo, 1 25 

* Price’s Handbook on Sanitation.nmo, 1 50 

Richards’s Cost of Food. A Study in Dietaries.nmo, 1 00 

Cost of Living as Modified by Sanitaiy Science.nmo, 1 00 

Richards and Woodman’s Air, Water, and Food from a Sanitary Stand¬ 
point. 8vo, 2 00 

* Richards and Williams’s The Dietary Computer.8vo, 1 50 

Rideal’s Sewage and Bacterial Purification of Sewage.8vo, 3 50 

Turneaure and Russell’s Public Water-supplies.8vo, 5 00 

Von Behring’s Suppression of Tuberculosis. (Bolduan.).nmo, 1 00 

Whipple’s Microscopy of Drinking-water.8vo, 3 50 

Woodhull’s Notes on Military Hygiene.i6mo, 1 50 

MISCELLANEOUS. 

De Fursac’s Manual of Psychiatry. (Rosanoff and Collins.)... .Large nmo, 2 50 
Emmons’s Geological Guide-book of the Rocky Mountain Excursion of the 

International Congress of Geologists.Large 8vo, 1 50 

Ferrel’s Popular Treatise on the Winds.8vo. 4 00 

Haines’s American Railway Management.nmo, 2 50 

Mott’s Composition, Digestibility, and Nutritive Value of Food. Mounted chart, 1 25 

Fallacy of the Present Theory of Sound.i6mo, 1 00 

Ricketts’s History of Rensselaer Polytechnic Institute, 1824-1894. .Small 8vo, 3 00 

Rostoski’s ( rum Diagnosis. (Bolduan.).nmo, 1 00 

Rotherham s Emphasized New Testament.Large 8vo, 2 00 

Steel’s Trea f! se on the Diseases of the Dog.8vo, 3 30 

Totten’s Important Question in Metrology.8vo, 2 50 

The World’s Columbian Exposition of 1893.4to, 1 00 

Von Behring’s Suppression of Tuberculosis. (Bolduan.).nmo, 1 00 

Winslow’s Elements of Applied Microscopy.nmo, x 50 


Worcester and Atkinson. Small Hospitals, Establishment and Maintenance; 

Su gestions for Hospital Architecture: Plans for Small Hospital, nmo, 1 25 

HEBREW AND CHALDEE TEXT-BOOKS. 

Green’s Elementary Hebrew Grammar.i2mo, 1 25 

Hebrew Chrestomathy.8vo, 2 00 

Gesenius’s Hebrew and Chaldee Lexicon to the Old Testament Scriptures. 

(Tregelles.).Small 4to, half morocco, 5 00 

Lettepis’s Hebrew Bible...8vo, 2 23 


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